RD Sharma Volume 1 - Determinants Solution For Class 12 Mathematics

The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

M₁₁ = –1

M₂₁ = 20

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × –1

= –1

C₂₁ = (–1)²⁺¹ × M₂₁

= 20 × –1

= –20

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁

= 5× (–1) + 0 × (–20)

= –5

Question 2.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Answer:

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

M₁₁ = 3

M₂₁ = 4

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 3

= 3

C₂₁ = (–1)²⁺¹ × 4

= –1 × 4

= –4

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁

= –1× 3 + 2 × (–4)

= –11

Question 3.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Answer:

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

M₁₁ = –1×2 – 5×2

M₁₁ = –12

M₂₁ = –3×2 – 5×2

M₂₁ = –16

M₃₁ = –3×2 – (–1) × 2

M₃₁ = –4

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × –12

= –12

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × –16

= 16

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –4

= –4

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (–12) + 4 × 16 + 3× (–4)

= –12 + 64 –12

= 40

Question 4.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Answer:

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

M₁₁ = b × ab – c × ca

M₁₁ = ab² – ac²

M₂₁ = a × ab – c × bc

M₂₁ = a²b – c²b

M₃₁ = a × ca – b × bc

M₃₁ = a²c – b²c

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (ab² – ac²)

= ab² – ac²

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × (a²b – c²b)

= c²b – a²b

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (a²c – b²c)

= a²c – b²c

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (ab² – ac²) + 1 × (c²b – a²b) + 1× (a²c – b²c)

= ab² – ac² + c²b – a²b + a²c – b²c

Question 5.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Answer:

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

M₁₁ = 5×1 – 7×0

M₁₁ = 5

M₂₁ = 2×1 – 7×6

M₂₁ = –40

M₃₁ = 2×0 – 5×6

M₃₁ = –30

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 5

= 5

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × –40

= 40

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –30

= –30

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 0× 5 + 1 × 40 + 3× (–30)

= 0 + 40 – 90

= 50

Question 6.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Answer:

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

M₁₁ = b × c – f × f

M₁₁ = bc– f²

M₂₁ = h × c – f × g

M₂₁ = hc – fg

M₃₁ = h × f – b × g

M₃₁ = hf – bg

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (bc– f²)

= bc– f²

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × (hc – fg)

= fg – hc

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (hf – bg)

= hf – bg

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= a× (bc– f²) + h× (fg – hc) + g× (hf – bg)

= abc– af² + hgf – h²c +ghf – bg²

Question 7.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Answer:

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

M₁₁ = 0(–1×0 – 5×1) – 1(1×0 – (–1)×1) + (–2)(1×5 – (–1)×(–1))

M₁₁ = – 9

M₂₁ = –1(–1×0 – 5×1) – 0(1×0 – (–1)×1) + 1(1×5 – (–1)×(–1))

M₂₁ = 9

M₃₁ = –1(1×0 – 5×(–2)) – 0(0×0 – (–1)×(–2)) + 1(0×5 – (–1)×1)

M₃₁ = –9

M₄₁ = –1(1×1 – (–1)×(–2)) – 0(0×1 – 1×(–2)) + 1(0×(–1) – 1×1)

M₄₁ = 0

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (–9)

= –9

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × 9

= –9

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –9

= –9

C₄₁ = (–1)⁴⁺¹ × M₄₁

= –1 × 0

= 0

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ + a₄₁× C₄₁

= 2× (–9) + (–3) × –9 + 1× (–9) + 2× 0

= – 18 + 27 –9

= 0

Question 8.

Evaluate the following determinants:

i.

ii.

iii.

iv.

Answer:

⇒ |A| = x(5x + 1) – (–7)x

|A| = 5x² + 8x

II.

⇒ |A| = cosθ × cosθ – (–sinθ) x sinθ

|A| = cos²θ + sin²θ

|A| = 1

III.

⇒ |A| = cos15° × cos75° + sin15° x sin75°

|A| = cos(75 – 15)°

|A| = cos60°

|A| = 0.5.

IV.

⇒ |A| = (a + ib)( a – ib) – (c + id)( –c + id)

= (a + ib)( a – ib) + (c + id)( c – id)

= a² – i² b² + c² – i² d²

= a² – (–1)b² + c² – (–1)d²

= a² + b² + c² + d²

Question 9.

Evaluate

Answer:

Since |AB|= |A||B|

= 2(17×12 – 5×20) – 3(13×12 – 5×15) + 7(13×20 – 15×17)

= 2(204 – 100) – 3(156 – 75) + 7(260 – 255)

= 2×104 – 3×81 + 7×5

= 208 – 243 +35

= 0

Now |A|² = |A|×|A|

|A|²= 0

Question 10.

Show that

Answer:

Using sin(A+B) = sinA × cosB + cosA × sinB

⇒ |A| = sin10° × cos80° + cos10° x sin80°

|A| = sin(10 + 80)°

|A| = sin90°

|A| = 1

Hence Proved

Question 11.

Evaluate by two methods.

Answer:

I. Expanding along the first row

= 2(1×1 – 4×(–2)) – 3(7×1 – (–2)×(–3)) – 5(7×4 – 1×(–3))

= 2(1 + 8) – 3(7 – 6) – 5(28 + 3)

= 2×9 – 3×1 – 5×31

= 18 – 3 – 155

= –140

II. Expanding along the second column

= 2(1×1 – 4×(–2)) – 7(3×1 – 4×(–5)) – 3(3×(–2) – 1×(–5))

= 2(1 + 8) – 7(3 + 20) – 3(–6 + 5)

= 2×9 – 7×23 – 3×(–1)

= 18 – 161 +3

= –140

Question 12.

Evaluate :

Answer:

Expanding along the first row

⇒ |A| = 0(0 – sinβ(–sinβ) ) –sinα(–sinα× 0 – sinβ cosα ) – cosα((–sinα)(–sinβ) – 0× cosα )

|A| = 0 + sinα sinβ cosα – cosα sinα sinβ

|A| = 0

Question 13.

Evaluate :

Answer:

Expanding along the second row

⇒ |A| = sinβ (cosα×cosα sinβ + sinα × sinα sinβ) + cosβ (cosα cosβ × cosα + sinα×sinα cosβ) – 0

|A| = sin² β (cos²α + sin²α) + cos²β (cos²α + sin²α)

|A| = sin² β (1) + cos²β (1)

|A| = sin² β + cos²β

|A| = 1

Question 14.

If and , verify that |AB| = |A| |B|.

Answer:

Now |A|= 2 × 1 – 2 × 5

|A|= 2 – 10

|A|= –8

Now |B|= 4 × 5 – 2 × (–3)

|B|= 20 + 6

|B|= 26

⇒ |A|×|B|= –8 × 26

|A|×|B|= –208

Now

|AB| = 18 × (–1) – 19 × 10

|AB| = –18 – 190

|AB| = –208

Hence |AB| =|A|×|B|.

Question 15.

If , then show that .

Answer:

Expanding along the first row

= 1(1×4 – 2×0) – 0(0×4 – 0×2) + 1(0×0 – 0×1)

= 1(4 – 0) + 0 + 1(0 + 0)

= 1×4

= 4

Now

Expanding along the first row

= 3(3×12 – 6×0) – 0(0×12 – 0×6) + 3(0×0 – 0×3)

= 3(36 – 0) + 0 + 3(0 + 0)

= 3×36

= 108

= 27 × 4

= 27 |A|

Hence, |3A|= 27 |A|

Question 16.

Find the value of x, if

Answer:

⇒ 2 × 1 – 4 × 5 = 2x × x – 4 × 6

⇒ 2 – 20 = 2x² – 24

⇒ 2x² = –18 +24

⇒ 2x² = 6

⇒ x² = 3

⇒ x = ±√ 3

Question 17.

Find the value of x, if

Answer:

⇒ 2 × 5 – 4 × 3 = x × 5 – 2x × 3

⇒ 10 – 12 = 5x – 6x

⇒ –x = –2

⇒ x = 2

Question 18.

Find the value of x, if

Answer:

⇒ 3 × 1 – x × x = 3 × 1 – 4 × 2

⇒ 3 – x² = 3 – 8

⇒ – x² = –5 –3

⇒ –x² = –8

⇒ x = ±2√ 2

Question 19.

Find the value of x, if

Answer:

⇒ 3x × 4 – 7 × 2 = 10

⇒ 12x – 14 = 10

⇒ 12x= 10 +14

⇒ 12x = 24

⇒ x = 2

Question 20.

Find the value of x, if

Answer:

⇒ (x+1)(x+2) – (x–1)(x–3) = 4 × 3 – 1 × (–1)

⇒ (x² + 2x + x + 2) – (x² – 3x – x + 3) = 12 + 1

⇒ –2x –1= 13

⇒ –2x = 14

⇒ x = –7

Question 21.

Find the value of x, if

Answer:

⇒ 2x × x – 5 × 8 = 6 × 3 – 5 × 8

⇒ 2x² – 40 = 18 – 40

⇒ 2x² = 18

⇒ x² = 9

⇒ x = ±3

Question 22.

Find the integral value of x, if

Answer:

Expanding along the first row

= x²(2×4 – 1×1) – x(0×4 – 1×3) + 1(0×1 – 2×3)

= x²(8 – 1) – x(0 – 3) + 1(0 – 6)

= 7x² + 3x –6

Also |A| = 28

⇒ 7x² + 3x – 6 =28

⇒ 7x² + 3x – 34 = 0

⇒ 7x² + 17x – 14x – 34 = 0

⇒ x(7x+ 17) – 2(7x +17) = 0

⇒ (x–2)(7x +17) = 0

Integer value of x is 2.

Question 23.

For what value of x matrix A is singular?

Answer:

⇒ (1 + x) × 8 – 7 × (3 – x) = 0

⇒ 8 + 8x – 21 + 7x = 0

⇒ 15x – 13 = 0

Question 24.

For what value of x matrix A is singular?

Answer:

Expanding along the first row

= (x–1) ((x–1) (x–1)– 1×1) – 1((x–1) – 1×1) + 1(1×1 – 1×(x–1))

= (x–1) (x² – 2x + 1 – 1) – 1(x–1 – 1) + 1(1 – x+1)

= x(x–1) (x– 2) – 1(x–2) – (x–2)

= (x– 2) {x(x–1) – 1 – 1}

= (x– 2) (x² – x – 2)

For singular |A| = 0,

(x– 2) (x² – x – 2) = 0

(x– 2) (x² – 2x + x – 2) = 0

(x–2)(x–2)(x+1) = 0

∴ x = –1 or 2

Also |A| = 28

⇒ 7x² + 3x – 6 =28

⇒ 7x² + 3x – 34 = 0

⇒ 7x² + 17x – 14x – 34 = 0

⇒ x(7x+ 17) – 2(7x +17) = 0

⇒ (x–2)(7x +17) = 0

Exercise 6.2

Question 1.

Evaluate the following determinant:

Answer:

Applying, R₂→ R₂ – R₁, we get,

So,

Question 2.

Evaluate the following determinant:

Answer:

Applying, C₁ → C₁ – 4 C₃, we get,

Applying, R₁ → R₁ + R₂ and R₃→ R₃ – R₂, we get

Now, applying R₂ → R₂ + 3 R₁, we get,

= 1[(109)(12) – (119)(11)] = 1308 – 1309

= – 1

So, Δ = – 1

Question 3.

Evaluate the following determinant:

Answer:

= a(bc – f²) – h(hc – fg) + g(hf – bg)

= abc – af² – ch² + fgh + fgh – bg²

= abc + 2fgh – af² – bg² – ch²

So, Δ = abc + 2fgh – af² – bg² – ch²

Question 4.

Evaluate the following determinant:

Answer:

Applying, R₂ → R₂ – R₁ and R₃ → R₃ – R_1, we get

= 2[1(24 – 4)] = 40

So, Δ = 40

Question 5.

Evaluate the following determinant:

Answer:

Applying C₃→ C₃ – C₂, we get,

Applying C₂ → C₂ + C₁, we get,

Applying C₂→ C₂ – 5C₁ and C₃ →C₃ – 5C₁ we get,

= 1[( – 7)( – 36) – ( – 20)( – 13)] = 252 – 260

= – 8

So, Δ = – 8

Question 6.

Evaluate the following determinant:

Answer:

Applying, R₁ → R₁ – 3R₂ and R₃ → R₃ + 5R₂ we get,

So, Δ = 0

Question 7.

Evaluate the following determinant:

Answer:

Applying C₁ → C₁ + C₂ + C₃ + C₄, we get,

Now, applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, R₄ → R₄ – R₁, we get

Now, applying R₁ → R₁ + 3R₃

= (40)(8)(40)(40) = 512000

So, Δ = 512000

Question 8.

Evaluate the following determinant:

Answer:

Applying R₃ → R₃ – R₁, we get,

So, Δ = 0

Question 9.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying R₃ → R₃ – R₂, we get

Applying R₂ → R₂ – R₁, we get

As, R₂ = R₃, therefore the value of the determinant is zero.

Question 10.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Taking ( – 2) common from C₁ we get,

As, C₁ = C₂, hence the value of the determinant is zero.

Question 11.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying C₃ → C₃ – C₂, gives

As, R₁ = R₃, so value so determinant is zero.

Question 12.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Multiplying R₁, R₂ and R₃ with a, b and c respectively we get,

Taking, abc common from C₃ gives,

As, C₁ = C₃ hence value of determinant is zero.

Question 13.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying C₃ → C₃ – C₂, we get,

Applying C₂→C₂ – C₁ gives,

As, C₂ = C₃, so the value of the determinant is zero.

Question 14.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying R₂→R₂ – R₁ and R₃ → R₃ – R₁, we get,

Taking (b – a) and (c – a) common from R₂ and R₃ respectively,

= [(b – a)(c – a)][(c + a) – (b + a) – ( – b + c)]

= [(b – a)(c – a)][c + a + b – a – b – c]

= [(b – a)(c – a)][0] = 0

Question 15.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying, C₁→C₁ – 8C₃

As, C₁ = C₂ hence, the determinant is zero.

Question 16.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Multiplying C₁, C₂ and C₃ with z, y and x respectively we get,

Now, taking y, x and z common from R₁, R₂ and R₃ gives,

Applying C₂ → C₂ – C₃ gives,

As, C₁ = C₂, therefore determinant is zero.

Question 17.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying C₂→C₂ – 7C₃, we get

As, C₁ = C₂, hence determinant is zero.

Question 18.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying C₃→C₃ – C₂, and C₄→C₄ – C₁

Taking 3 common from C₄ we get,

As, C3 = C4 so, the determinant is zero.

Question 19.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying, C₂→C₂ + C₁ and C₃→C₃ + C₁

Taking 2 common from R₂ we get,

As, R₂ = R₃, hence value of determinant is zero.

Question 20.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying, C₁→C₁ – C₂, we get

As C₁ = C₃ hence determinant is zero.

Question 21.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Multiplying C₁ with sin δ, C₂ with cos δ, we get

Now, applying, C₂→C₂ – C₁, we get,

As C₂ = C₃ hence determinant is zero.

Question 22.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Applying C₁→C₁ + C₂, we get

Using, sin(90 – A) = cos A, sin² A + cos² A = 1 ,and cos 180˚ = – 1,

Taking, ( – 1) common from C₁, we get

Therefore, as C₁ = C₃ determinant is zero.

Question 23.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Multiplying R₂ with sin y and R₃ with cos y we get,

Now, applying R₂→R₂ + R₃, we get,

Taking ( – 1) common from R₂, we get

As R₁ = R₂ hence determinant is zero.

Question 24.

Without expanding, show that the value of each of the following determinants is zero:

Answer:

Multiplying C₂ with and C₃ with we get,

Taking common from C₂ and C₃ we get,

Applying C₂→C₂ + C₃

As C₁ = C₂ hence determinant is zero.

Question 25.

Without expanding, show that the value of each of the following determinants is zero:

, where A, B, C are the angles of ΔABC.

Answer:

Now,

Δ = sin² A (cot B – cot C) – cot A (sin² B – sin² C) + 1 (sin²B cot C – cot B sin² C

As A, B and C are angles of a triangle,

A + B + C = 180°

Δ = sin² A cot B – sin² A cot C – cot A sin² B + cot A sin²C + sin²B cot C – cot B sin² C

By using formulae,

Δ = 0

Hence, Proved.

Question 26.

Evaluate the following:

Answer:

Applying, C₂→C₂ + C₁

Taking, (a + b + c) common,

Applying R₂→R₂ – R₁, and R₃→R₃ – R₁

Taking, (b – c) and (c – a) common,

= (a + b + c)(b – a)(c – a)(b – c)

So, Δ = (a + b + c)(b – a)(c – a)(b – c)

Question 27.

Evaluate the following:

Answer:

Applying, R₂→R₂ – R₁ and R₃→R₃ – R₁ we get,

Taking (a – b) and (a – c) common we get,

= (a – b)(c – a)(b – c)

So, Δ = (a – b)(b – c)(c – a)

Question 28.

Evaluate the following:

Answer:

Applying, C₁→C₁ + C₂ + C₃, we have,

Taking, (3x + λ) common, we get

Applying, R₂→R₂ – R₁, R₃→R₃ – R₁, we get,

= λ²(3x + λ)

So, Δ = λ²(3x + λ)

Question 29.

Evaluate the following:

Answer:

Applying, C₁→C₁ + C₂ + C₃, we get,

Taking, (a + b + c) we get,

Applying, R₂→R₂ – R₁, R₃→R₃ – R₁, we get,

= (a + b + c)[(a – b)(a – c) – (b – c)(c – b)]

= (a + b + c)[a² – ac – ab + bc + b² + c² – 2bc]

= (a + b + c)[a² + b² + c² – ac – ab – bc]

So, Δ = (a + b + c)[a² + b² + c² – ac – ab – bc]

Question 30.

Evaluate the following:

Answer:

Applying, C₁→C₁ + C₂ + C₃, we get,

Applying, R₂→R₂ – R₁, R₃→R₃ – R₁, we get,

= (2 + x)(x – 1)²

So, Δ = (2 + x)(x – 1)²

Question 31.

Evaluate the following:

Answer:

= 0(0 – y³z³) – xy²(0 – x²yz³) + xz²(x²y³z – 0)

= 0 + x³y³z³ + x³y³z³

= 2x³y³z³

So, Δ = 2x³y³z³

Question 32.

Evaluate the following:

Answer:

Applying R₁→R₁ – R₂ and R₃→R₃ – R₂

Applying, C₂→C₂ – C₁

= a[a(a + x + y) + az] + 0 + 0

= a²(a + x + y + z)

So, Δ = a²(a + x + y + z)

Question 33.

If ,, then prove that Δ + Δ₁ = 0.

Answer:

As |A| = |A|^T

If any two rows or columns of the determinant are interchanged, then determinant changes its sign

So, Δ = 0

Question 34.

Prove the following identities:

Answer:

L.H.S =

Apply C₁→C₁ + C₂ + C₃

Taking (a + b + c) common from C₁ we get,

Applying, R₃→R₃ – 2R₁

= (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)]

= a³ + b³ + c³ – 3abc

As, L.H.S = R.H.S

Hence, proved.

Question 35.

Prove the following identities:

Answer:

L.H.S =

As |A| = |A|^T

So,

If any two rows or columns of the determinant are interchanged, then determinant changes its sign

Apply C₁→C₁ + C₂ + C₃

Taking (a + b + c) common from C₁ we get,

Applying, R₃→R₃ – 2R₁

= – (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)]

= 3abc – a³ – b³ – c³

As, L.H.S = R.H.S, hence proved.

Question 36.

Prove the following identities:

Answer:

L.H.S =

Applying, C₁→C₁ + C₂ + C₃

Apply, C₂→C₂ – C₁, and C₃→C₃ – C₁, we have

Hence, proved.

Question 37.

Prove the following identities:

Answer:

R.H.S = 2(a + b + c)²

Applying C₁→C₁ + C₂ + C₃, we have

Taking, 2(a + b + c) common we get,

Now, applying R₂→R₂ – R₁ and R₃→R₃ – R₁, we get,

Thus, we have

L.H.S = 2(a + b + c)[1(a + b + c)²]

= 2(a + b + c)³ = R.H.S

Hence, proved.

Question 38.

Prove the following identities:

Answer:

L.H.S =

Applying, R₁→R₁ + R₂ + R₃, we get,

Taking (a + b + c) common we get,

Applying C₂→C₂ – C₁ and C₃→C₃ – C₁, we get,

= (a + b + c)³ = R.H.S

Hence, proved.

Question 39.

Prove the following identities:

Answer:

Applying, R₂→R₂ – R₁ and R₃→R₃ – R₁, we get,

Applying R₃→R₃ – R₂, we get,

= (a – b)(a – c)(b – c) = R.H.S

Hence, proved.

Question 40.

Prove the following identities:

Answer:

Applying R₁→R₁ + R₂ + R₃, we get,

Taking, (3a + 2b) common we get,

Applying, C₁→C₁ – C₂ and C₃→C₃ – C₂, we get,

= 3(a + b)b²(3) = 9(a + b) b²

= R.H.S

Hence, proved.

Question 41.

Prove the following identities:

Answer:

Applying, R₁→a R₁, R₂→b R₂, R₃→c R₃

Hence, proved.

Question 42.

Prove the following identities:

=

Answer:

= – xyz(x – y)(z – y)[z² + y² + zy – x² – y² – xy]

= – xyz(x – y)(z – y)[(z – x)(z + x0 + y(z – x)]

= – xyz(x – y)(z – y)(z – x)(x + y + z)

= R.H.S

Hence, proved.

Question 43.

Prove the following identities:

Answer:

Applying, C₁→C₁ + C₂ – 2C₃

Taking (a² + b² + c²), common, we get,

Applying R₂→R₂ – R₁ and R₃→R₃ – R₁, we get,

= (a² + b² + c²)(b – a)(c – a)[(b + a)( – b) – ( – c)(c + a)]

= (a² + b² + c²)(a – b)(c – a)(b – c)(a + b + c)

= R.H.S

Hence, proved.

Question 44.

Prove the following identities:

Answer:

Applying, R₃→R₃ – R₂

Applying, R₂→R₂ – R₁

= [(2a + 4)(1) – (1)(2a + 6)]

= – 2

= R.H.S

Hence, proved.

Question 45.

Prove the following identities:

Answer:

Applying, C₂→C₂ – 2C₁ – 2C₃, we get,

Taking, – (a² + b² + c²) common from C₂ we get,

Applying R₂→R₂ – R₁ and R₃→R₃ – R₁, we get

= – (a² + b² + c²)(a – b)(c – a)[( – (b + a))( – b) – (c)(c + a)]

= (a – b)(b – c)(c – a)(a + b + c)(a² + b² + c²)

= R.H.S

Hence, proved.

Question 46.

Prove the following identities:

Answer:

Applying, R₂→R₂ – R₁, and R₃→R₃ – R₁

= (b – a)(c – a)[((b + a – c))(c² + a² + ac) – (b² + a² + ab)(c² + a² + ac)]

= – (a – b)(c – a)(b – c)(a² + b² + c²)

= R.H.S

Hence, proved.

Question 47.

Prove the following identities:

Answer:

Taking, a,b,c common from C₁, C₂, C₃ respectively we get,

Applying, C₁→C₁ + C₂ + C₃, we get,

Applying, C₂→C₂ – C₁ and C₃→C₃ – C₁, we get,

Applying, C₁→C₁ + C₂ + C₃, we get,

Taking c, a, b common from C₁, C₂, C₃ respectively, we get,

Applying, R₃→R₃ – R₁, we have

= 2a²b²c²(2)

= 4a²b²c² = R.H.S

Hence, proved.

Question 48.

Prove the following identities:

Answer:

Applying, C₁→C₁ + C₂ + C₃, we get,

Taking (3x + 4) common we get,

Applying, R₂→R₂ – R₁ and R₃→R₃ – R₁, we get,

= 16(3x + 4)

Hence proved.

Question 49.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₂→ C₂ – pC₁, we get

Applying C₃→ C₃ – qC₁, we get

Applying C₃→ C₃ – pC₂, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = 1[(1)(7) – (3)(2)] – 0 + 0

∴ Δ = 7 – 6 = 1

Thus,

Question 50.

Prove the following identities –

= (a + b - c)(b + c - a)(c + a - b)

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₁→ R₁ – R₃, we get

Taking the term (a – b – c) common from R₁, we get

Applying C₂→ C₂ + C₁, we get

Applying C₃→ C₃ + C₁, we get

Expanding the determinant along R₁, we have

Δ = (a – b – c)[–1(b + a – c)(c + a – b) – 0 + 0]

⇒ Δ = –(a – b – c)(b + a – c)(c + a – b)

∴ Δ = (b + c – a)(a + b – c)(c + a – b)

Thus,

Question 51.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (a² + b² + 2ab) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (a + b)² [(a² – b²)(a² – 2ab) – (b² – 2ab)(2ab – b²)]

⇒ Δ = (a + b)² [a⁴ – 2a³b – b²a² + 2ab³ – 2ab³ + b⁴ + 4a²b² – 2ab³]

⇒ Δ = (a + b)² [a⁴ – 2a³b + 3a²b² – 2ab³ + b⁴]

⇒ Δ = (a + b)² [a⁴ + b⁴ + 2a²b² – 2a³b – 2ab³ + a²b²]

⇒ Δ = (a + b)² [(a² + b²)² – 2ab(a² + b²) + (ab)²]

⇒ Δ = (a + b)² [(a² + b² – ab)²]

⇒ Δ = [(a + b)(a² + b² – ab)]²

∴ Δ = (a³ + b³)²

Thus,

Question 52.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (a² + a + 1) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (a² + a + 1)(1)[(1 – a²)(1 – a) – (a² – a)(a – a²)]

⇒ Δ = (a² + a + 1)(1 – a – a² + a³ – a³ + a⁴ + a² – a³)

⇒ Δ = (a² + a + 1)(1 – a – a³ + a⁴)

⇒ Δ = (a² + a + 1)(a⁴ – a³ – a + 1)

⇒ Δ = (a² + a + 1)[a³(a – 1) – (a – 1)]

⇒ Δ = (a² + a + 1)(a – 1)(a³ – 1)

⇒ Δ = (a – 1)(a² + a + 1)(a³ – 1)

⇒ Δ = (a³ – 1)(a³ – 1)

∴ Δ = (a³ – 1)²

Thus,

Question 53.

Prove the following identities –

= 2(a + b)(b + c)(c + a)

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Applying C₂→ C₂ + C₁, we get

Applying C₃→ C₃ + C₁, we get

Taking (a + b) and (c + a) common from C₂ and C₃, we get

Applying R₂→ R₂ + R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₃, we have

Δ = (a + b)(c + a)[(–c + a)(–2) – (–b – a)(2)]

⇒ Δ = (a + b)(c + a)[2c – 2a + 2a + 2b]

⇒ Δ = (a + b)(c + a)(2b + 2c)

∴ Δ = 2(a + b)(b + c)(c + a)

Thus,

Question 54.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₁→ R₁ – R₃, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

⇒ Δ = 0 + (2c)[(b)(a + b – c)] + (–2b)[–(c)(c + a – b)]

⇒ Δ = 2bc(a + b – c) + 2bc(c + a – b)

⇒ Δ = 2bc[(a + b – c) + (c + a – b)]

⇒ Δ = 2bc[2a]

∴ Δ = 4abc

Thus,

Question 55.

Prove the following identities –

Answer:

Let

Multiplying a, b and c to R₁, R₂ and R₃, we get

Dividing C₁, C₂ and C₃ with a, b and c, we get

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₁→ R₁ – R₃, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = 0 + (2c²)[(b²)(a² + b² – c²)] + (–2b²)[–(c²)(c² + a² – b²)]

⇒ Δ = 2b²c²(a² + b² – c²) + 2b²c²(c² + a² – b²)

⇒ Δ = 2b²c² [(a² + b² – c²) + (c² + a² – b²)]

⇒ Δ = 2b²c²[2a²]

∴ Δ = 4a²b²c²

Thus,

Question 56.

Prove the following identities –

Answer:

Let

Taking a², b² and c² common from C₁, C₂ and C₃, we get

Taking a, b and c common from R₁, R₂ and R₃, we get

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₂→ C₂ – C₃, we get

Expanding the determinant along R₁, we have

Δ = (a³b³c³)[0 – 0 + 1(1)(1) – (1)(–1)]

⇒ Δ = (a³b³c³)[1 + 1]

∴ Δ = 2a³b³c³

Thus,

Question 57.

Prove the following identities –

Answer:

Let

Multiplying c, a and b to R₁, R₂ and R₃, we get

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₁→ R₁ – R₃, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

∴ Δ = 4abc

Thus,

Question 58.

Prove the following identities –

Answer:

Let

Multiplying a, b and c to R₁, R₂ and R₃, we get

Dividing C₁, C₂ and C₃ with a, b and c, we get

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (a – b – c) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (ab + bc + ca)(1)[(ab + bc + ca)(ab + bc + ca)]

∴ Δ = (ab + bc + ca)³

Thus,

Question 59.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (5x + λ) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (5x + λ)[(1)(λ – x)(λ – x)]

∴ Δ = (5x + λ)(λ – x)²

Thus,

Question 60.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (5x + 4) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (5x + 4)[(1)(4 – x)(4 – x)]

∴ Δ = (5x + 4)(4 – x)²

Thus,

Question 61.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₁→ R₁ – R₃, we get

Taking the term (–2x) common from R₁, we get

Applying C₂→ C₂ – C₃, we get

Expanding the determinant along R₁, we have

Δ = (–2x)[(z)(–y) – (y)(z)]

⇒ Δ = (–2x)(–yz –yz)

⇒ Δ = (–2x)(–2yz)

∴ Δ = 4xyz

Thus,

Question 62.

Prove the following identities –

= abc(a² + b² + c²)³

Answer:

Let

Taking a, b and c common from C₁, C₂ and C₃, we get

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Taking the term (a² + b² + c²) common from R₁, we get

Applying R₂→ R₂ – R₃, we get

Taking the term (a² + b² + c²) common from R₂, we get

Applying C₂→ C₂ + C₁, we get

Expanding the determinant along R₁, we have

Δ = (abc)(a² + b² + c²)²(–1)[(a² + b² – c²) – (2b² + 2a²)]

⇒ Δ = (abc)(a² + b² + c²)²[–(a² + b² – c²) + (2b² + 2a²)]

⇒ Δ = (abc)(a² + b² + c²)²[–a² – b² + c² + 2b² + 2a²]

⇒ Δ = (abc)(a² + b² + c²)²[a² + b² + c²]

∴ Δ = (abc)(a² + b² + c²)³

Thus,

Question 63.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (3 + a) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (3 + a)(1)[(a)(a) – 0]

⇒ Δ = (3 + a)(a²)

∴ Δ = a³ + 3a²

Thus,

Question 64.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (x + y + z) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (x + y + z)(1)[0 – (–(x + y + z)(x + y + z))]

⇒ Δ = (x + y + z)(x + y + z)(x + y + z)

∴ Δ = (x + y + z)³

Thus,

Question 65.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term (x + y + z) common from R₁, we get

Applying C₁→ C₁ – C₂, we get

Applying C₁→ C₁ – C₃, we get

Expanding the determinant along C₁, we have

Δ = (x + y + z)(x – z)[(1)(x) – (z)(1)]

⇒ Δ = (x + y + z)(x – z)(x – z)

∴ Δ = (x + y + z)(x – z)²

Thus,

Question 66.

Prove the following identities –

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₁→ C₁ + C₂, we get

Applying C₁→ C₁ + C₃, we get

Taking the term (a + x + y + z) common from C₁, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₁, we have

Δ = (a + x + y + z)(1)[(a)(a) – (0)(0)]

⇒ Δ = (a + x + y + z)(a)(a)

∴ Δ = a²(a + x + y + z)

Thus,

Question 67.

Prove the following identities –

= 2(a - b)(b - c)(c - a)(a + b + c)

Answer:

Let

Taking 2 common from C₂, we get

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

We have the identity a³ – b³ = (a – b)(a² + ab + b²)

Taking (b – a) and (c – a) common from R₂ and R₃, we get

We know that the sign of a determinant changes if any two rows or columns are interchanged.

By interchanging C₁ and C₂, we get

Expanding the determinant along C₁, we have

Δ = –2(b – a)(c – a)(1)[(b² + ba + a²) – (c² + ca + a²)]

⇒ Δ = 2(a – b)(c – a)[b² + ba + a² – c² – ca – a²]

⇒ Δ = 2(a – b)(c – a)[b² + ba – c² – ca]

⇒ Δ = 2(a – b)(c – a)[b² – c² + (ba – ca)]

⇒ Δ = 2(a – b)(c – a)[(b – c)(b + c) + (b – c)a]

⇒ Δ = 2(a – b)(c – a)(b – c)(b + c + a)

∴ Δ = 2(a – b)(b – c)(c – a)(a + b + c)

Thus,

Question 68.

Without expanding, prove that .

Answer:

Let

We know that the sign of a determinant changes if any two rows or columns are interchanged.

By interchanging R₁ and R₂, we get

By interchanging R₂ and R₃, we get

Hence,

Let us once again consider

By interchanging R₁ and R₂, we get

By interchanging C₁ and C₂, we get

Recall that the value of a determinant remains same if it its rows and columns are interchanged.

Hence,

Thus,

Question 69.

Show that where a,b,c are in A.P.

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₃, we get

Given that a, b and c are in an A.P. Using the definition of an arithmetic progression, we have

b – a = c – b

⇒ b + b = c + a

⇒ 2b = c + a

∴ a + c = 2b

By substituting this in the above equation to find Δ, we get

Taking 2 common from R₁, we get

Applying R₁→ R₁ – R₂, we get

∴ Δ = 0

Thus, when a, b and c are in A.P.

Question 70.

Show that where α, β and γ are in A.P.

Answer:

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₃, we get

Given that α, β and γ are in an A.P. Using the definition of an arithmetic progression, we have

β – α = γ – β

⇒ β + β = γ + α

⇒ 2β = γ + α

∴ α + γ = 2β

By substituting this in the above equation to find Δ, we get

Taking 2 common from R₁, we get

Applying R₁→ R₁ – R₂, we get

∴ Δ = 0

Thus, when α, β and γ are in A.P.

Question 71.

If a, b, c are real numbers such that , then show that either a + b + c = 0 or a = b = c.

Answer:

Let

Given that Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Taking the term 2(a + b + c) common from R₁, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = 2(a + b + c)(1)[(b – c)(c – b) – (c – a)(b – a)]

⇒ Δ = 2(a + b + c)(bc – b² – c² + cb – cb + ca + ab – a²)

∴ Δ = 2(a + b + c)(ab + bc + ca – a² – b² – c²)

We have Δ = 0

⇒ 2(a + b + c)(ab + bc + ca – a² – b² – c²) = 0

⇒ (a + b + c)(ab + bc + ca – a² – b² – c²) = 0

Case – I:

a + b + c = 0

Case – II:

ab + bc + ca – a² – b² – c² = 0

⇒ a² + b² + c² – ab – bc – ca = 0

Multiplying 2 on both sides, we have

2(a² + b² + c² – ab – bc – ca) = 0

⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

⇒ a² – 2ab + b² + b² – 2bc + c² + c² – 2ca + a² = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0

We know (a – b)² ≥ 0, (b – c)² ≥ 0, (c – a)² ≥ 0

If the sum of three non-negative numbers is zero, then each of the numbers is zero.

⇒ (a – b)² = 0 = (b – c)² = (c – a)²

⇒ a – b = 0 = b – c = c – a

⇒ a = b = c

Thus, if, then either a + b + c = 0 or a = b = c.

Question 72.

If , find the value of .

Answer:

Let

Given that Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₂→ R₂ – R₃, we get

Expanding the determinant along R₁, we have

Δ = (p – a)[(q – b)(r) – (b)(c – r)] – (b – q)[–a(c – r)]

⇒ Δ = r(p – a)(q – b) – b(p – a)(c – r) + a(b – q)(c – r)

∴ Δ = r(p – a)(q – b) + b(p – a)(r – c) + a(q – b)(r – c)

We have Δ = 0

⇒ r(p – a)(q – b) + b(p – a)(r – c) + a(q – b)(r – c) = 0

On dividing the equation with (p – a)(q – b)(r – c), we get

Thus,

Question 73.

Show that x = 2 is a root of the equation and solve it completely.

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Taking the term (x – 2) common from R₂, we get

Applying R₃→ R₃ – R₁, we get

Taking the term (x + 3) common from R₃, we get

Applying C₁→ C₁ + C₃, we get

Expanding the determinant along C₁, we have

Δ = (x – 2)(x + 3)(x – 1)[(–3)(1) – (2)(1)]

⇒ Δ = (x – 2)(x + 3)(x – 1)(–5)

∴ Δ = –5(x – 2)(x + 3)(x – 1)

The given equation is Δ = 0.

⇒ –5(x – 2)(x + 3)(x – 1) = 0

⇒ (x – 2)(x + 3)(x – 1) = 0

Case – I:

x – 2 = 0 ⇒ x = 2

Case – II:

x + 2 = 0 ⇒ x = –3

Case – III:

x – 1 = 0 ⇒ x = 1

Thus, 2 is a root of the equation and its other roots are –3 and 1.

Question 74.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₁→ C₁ + C₂, we get

Applying C₁→ C₁ + C₃, we get

Taking the term (x + a + b + c) common from C₁, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₁, we have

Δ = (x + a + b + c)(1)[(x)(x) – (0)(0)]

⇒ Δ = (x + a + b + c)(x)(x)

∴ Δ = x²(x + a + b + c)

The given equation is Δ = 0.

⇒ x²(x + a + b + c) = 0

Case – I:

x² = 0 ⇒ x = 0

Case – II:

x + a + b + c = 0 ⇒ x = –(a + b + c)

Thus, 0 and –(a + b + c) are the roots of the given determinant equation.

Question 75.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₁→ C₁ + C₂, we get

Applying C₁→ C₁ + C₃, we get

Taking the term (3x + a) common from C₁, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₁, we have

Δ = (3x + a)(1)[(a)(a) – (0)(0)]

⇒ Δ = (3x + a)(a)(a)

∴ Δ = a²(3x + a)

The given equation is Δ = 0.

⇒ a²(3x + a) = 0

However, a ≠ 0 according to the given condition.

⇒ 3x + a = 0

⇒ 3x = –a

Thus, is the root of the given determinant equation.

Question 76.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₁→ C₁ + C₂, we get

Applying C₁→ C₁ + C₃, we get

Taking the term (3x – 2) common from C₁, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₁, we have

Δ = (3x – 2)(1)[(3x – 11)(3x – 11) – (0)(0)]

⇒ Δ = (3x – 2)(3x – 11)(3x – 11)

∴ Δ = (3x – 2)(3x – 11)²

The given equation is Δ = 0.

⇒ (3x – 2)(3x – 11)² = 0

Case – I:

3x – 2 = 0

⇒ 3x = 2

Case – II:

(3x – 11)² = 0

⇒ 3x – 11 = 0

⇒ 3x = 11

Thus, and are the roots of the given determinant equation.

Question 77.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Taking (a – x) and (b – x) common from R₂ and R₃, we get

Expanding the determinant along C₁, we have

Δ = (a – x)(b – x)(1)[(1)(b + x) – (1)(a + x)]

⇒ Δ = (a – x)(b – x)[b + x – a – x]

∴ Δ = (a – x)(b – x)(b – a)

The given equation is Δ = 0.

⇒ (a – x)(b – x)(b – a) = 0

However, a ≠ b according to the given condition.

⇒ (a – x)(b – x) = 0

Case – I:

a – x = 0 ⇒ x = a

Case – II:

b – x = 0 ⇒ x = b

Thus, a and b are the roots of the given determinant equation.

Question 78.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₁→ C₁ + C₂, we get

Applying C₁→ C₁ + C₃, we get

Taking the term (x + 9) common from C₁, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₁, we have

Δ = (x + 9)(1)[(x – 1)(x – 1) – (0)(0)]

⇒ Δ = (x + 9)(x – 1)(x – 1)

∴ Δ = (x + 9)(x – 1)²

The given equation is Δ = 0.

⇒ x²(x + a + b + c) = 0

Case – I:

x+ 9 = 0 ⇒ x = –9

Case – II:

(x – 1)² = 0

⇒ x – 1 = 0

∴ x = 1

Thus, –9 and 1 are the roots of the given determinant equation.

Question 79.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Taking (b – x) and (c – x) common from R₂ and R₃, we get

Expanding the determinant along C₁, we have

Δ = (b – x)(c – x)(1)[(1)(c² + cx + x²) – (1)(b² + bx + x²)]

⇒ Δ = (b – x)(c – x)[c² + cx + x² – b² – bx – x²]

⇒ Δ = (b – x)(c – x)[c² – b² + cx – bx]

⇒ Δ = (b – x)(c – x)[(c – b)(c + b) + (c – b)x]

∴ Δ = (b – x)(c – x)(c – b)(c + b + x)

The given equation is Δ = 0.

⇒ (b – x)(c – x)(c – b)(c + b + x) = 0

However, b ≠ c according to the given condition.

⇒ (b – x)(c – x)(c + b + x) = 0

Case – I:

b – x = 0 ⇒ x = b

Case – II:

c – x = 0 ⇒ x = c

Case – III:

c + b + x = 0 ⇒ x = –(b + c)

Thus, b, c and –(b + c) are the roots of the given determinant equation.

Question 80.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₃, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₂, we have

Δ = – (1)[(–4 – x)(3) – (6)(x – 8)]

⇒ Δ = – [–12 – 3x – 6x + 48]

⇒ Δ = – [– 9x + 36]

∴ Δ = 9x – 36

The given equation is Δ = 0.

⇒ 9x – 36 = 0

⇒ 9x = 36

∴ x = 4

Thus, 4 is the root of the given determinant equation.

Question 81.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Taking the term p common from R₂, we get

Applying C₁→ C₁ – C₂, we get

Expanding the determinant along C₁, we have

Δ = p(2 – x)[(1)(1) – (1)(x)]

∴ Δ = p(2 – x)(1 – x)

The given equation is Δ = 0.

⇒ p(2 – x)(1 – x) = 0

Assuming p ≠ 0, we get

⇒ (2 – x)(1 – x) = 0

Case – I:

2 – x = 0 ⇒ x = 2

Case – II:

1 – x = 0 ⇒ x = 1

Thus, 1 and 2 are the roots of the given determinant equation.

Question 82.

Solve the following determinant equations:

Answer:

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₁→ C₁ + C₂, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – 3R₁, we get

Expanding the determinant along C₁, we have

Δ = (1)[(10)(2 – 3sin(3θ)) – (20)(cos(2θ) – sin(3θ))]

⇒ Δ = [20 – 30sin(3θ) – 20cos(2θ) + 20sin(3θ)]

⇒ Δ = 20 – 10sin(3θ) – 20cos(2θ)

From trigonometry, we have sin(3θ) = 3sinθ – 4sin³θ and cos(2θ) = 1 – 2sin²θ.

⇒ Δ = 20 – 10(3sinθ – 4sin³θ) – 20(1 – 2sin²θ)

⇒ Δ = 20 – 30sinθ + 40sin³θ – 20 + 40sin²θ

⇒ Δ = –30sinθ + 40sin²θ + 40sin³θ

∴ Δ = 10(sinθ)(–3 + 4sinθ + 4sin²θ)

The given equation is Δ = 0.

⇒ 10(sinθ)(–3 + 4sinθ + 4sin²θ) = 0

⇒ (sinθ)(–3 + 4sinθ + 4sin²θ) = 0

Case – I:

sin θ = 0 ⇒ θ = kπ, where k ϵ Z

Case – II:

–3 + 4sinθ + 4sin²θ = 0

⇒ 4sin²θ + 4sinθ – 3 = 0

⇒ 4sin²θ + 6sinθ – 2sinθ – 3 = 0

⇒ 2sinθ(2sinθ + 3) – 1(2sinθ + 3) = 0

⇒ (2sinθ – 1)(2sinθ + 3) = 0

⇒ 2sinθ – 1 = 0 or 2sinθ + 3 = 0

⇒ 2sinθ = 1 or 2sinθ = –3

However, as –1 ≤ sin θ ≤ 1.

, where k ϵ Z

Thus, kπ and for all integral values of k are the roots of the given determinant equation.

Question 83.

If a, b and c are all non-zero and , then prove that .

Answer:

Let

Given that Δ = 0.

We can write the determinant Δ as

Taking a, b and c common from C₁, C₂ and C₃, we get

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying C₁→ C₁ + C₂, we get

Applying C₁→ C₁ + C₃, we get

Taking common from C₁, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₁, we have

We have Δ = 0.

It is given that a, b and c are all non-zero.

Thus, when and a, b, c are all non-zero.

Question 84.

If , then using properties of determinants, find the value of , where .

Answer:

Let

Given that Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₃, we have

⇒ Δ = (c – z)[0 – (–x)(y)] – 0 + z[(a)(y) – (–x)(b – y)]

⇒ Δ = (c – z)(xy) + z[ay + xb – xy]

⇒ Δ = cxy – xyz + ayz + bxz – xyz

∴ Δ = ayz + bxz + cxy – 2xyz

We have Δ = 0

⇒ ayz + bxz + cxy – 2xyz = 0

⇒ ayz + bxz + cxy = 2xyz

Thus, when.

Exercise 6.3

Question 1.

Find the area of the triangle with vertices at the points:

(3, 8), (– 4, 2) and (5, – 1)

Answer:

Given: – Vertices of the triangle:

(3, 8), (– 4, 2) and (5, – 1)

We know that,

If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now, substituting given value in above formula

Expanding along R₁

sq.units

Thus area of triangle is sq.units

Question 2.

Find the area of the triangle with vertices at the points:

(2, 7) (1, 1) and (10, 8)

Answer:

Given: – Vertices of the triangle:

(2, 7) (1, 1) and (10, 8)

We know that,

If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now, substituting given value in above formula

Expanding along R₁

sq.units

Thus area of triangle is sq.units

Question 3.

Find the area of the triangle with vertices at the points:

(– 1, – 8), (– 2, – 3) and (3, 2)

Answer:

Given: – Vertices of the triangle:

(– 1, – 8), (– 2, – 3) and (3, 2)

We know that,

If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now, substituting given value in above formula

Expanding along R₁

sq.units

as area cannot be negative

Therefore, 15 sq.unit is the area

Thus area of triangle is 15 sq.units

Question 4.

Find the area of the triangle with vertices at the points:

(0, 0) (6, 0) and (4, 3)

Answer:

Given: – Vertices of the triangle:

(0, 0) (6, 0) and (4, 3)

We know that,

If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now, substituting given value in above formula

Expanding along R₁

= 9 sq.units

Thus area of triangle is 9 sq.units

Question 5.

Using determinants show that the following points are collinear:

(1, – 1), (2, 1) and (4, 5)

Answer:

Given: – (1, – 1), (2, 1) and (4, 5) are three points

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero

Now, we know that,

vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

R.H.S

Expanding along R₁

= 0

= LHS

Since, Area of triangle is zero.

Hence, points are collinear.

Question 6.

Using determinants show that the following points are collinear:

(3, – 2), (8, 8) and (5, 2)

Answer:

Given: – (3, – 2), (8, 8) and (5, 2) are three points

Tip: – For Three points to be collinear, the area of triangle formed by these points will be zero

Now, we know that,

vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

R.H.S

Expanding along R₁

= 0

= LHS

Since, Area of triangle is zero

Hence, points are collinear.

Question 7.

Using determinants show that the following points are collinear:

(2, 3), (– 1, – 2) and (5, 8)

Answer:

Given: – (2, 3), (– 1, – 2) and (5, 8) are three points

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero

Now, we know that,

vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

R.H.S

Expanding along R₁

= 0

= LHS

Since, Area of triangle is zero

Hence, points are collinear.

Question 8.

If the points (a, 0), (o, b) and (1, 1) are collinear, prove that a + b = ab.

Answer:

Given: – (a, 0), (o, b) and (1, 1) are collinear points

To Prove: – a + b = ab

Proof: –

Tip: – If Three points to be collinear, then the area of the triangle formed by these points will be zero

Now, we know that,

vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Thus

Expanding along R₁

⇒

⇒ a + b = ab

Hence Proved

Question 9.

Using determinants prove that the points (a, b) (a’, b’) and (a – a’, b – b’) are collinear if ab’ = a’b.

Answer:

Given: – (a, b) (a’, b’) and (a – a’, b – b’) are points and ab’ = a’b

To Prove: – (a, b) (a’, b’) and (a – a’, b – b’) are collinear points

Proof: –

Tip: – If three points to be collinear, then the area of the triangle formed by these points will be zero.

Now, we know that,

vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Thus

Expanding along R₁

⇒

⇒ ab^’ – a^’b = 0

⇒ ab^’ = a^’b

Hence, Proved.

Question 10.

Find the value of λ so that the points (1, – 5), (– 4, 5) and (λ, 7) are collinear.

Answer:

Given: – (1, – 5), (– 4, 5) and (λ, 7) are collinear

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero

Now, we know that,

vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

Expanding along R₁

⇒

⇒ 50 – 10λ = 0

⇒ λ = – 5

⇒

Question 11.

Find the value of x if the area of a triangle is 35 square cms with vertices (x, 4), (2, – 6) and (5, 4).

Answer:

Given: – Vertices of triangle are (x, 4), (2, – 6) and (5, 4) and area of triangle is 35 sq.cms

Tip: – If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

⇒

Removing modulus

⇒

Expanding along R₁

⇒

⇒ [x(– 10) – 4(– 3) + 1(8 – 30)] = ± 70

⇒ [ – 10x + 12 + 38] = ± 70

⇒ ±70 = – 10x + 50

Taking + ve sign, we get

⇒ + 70 = – 10x + 50

⇒ 10x = – 20

⇒ x = – 2

Taking – ve sign, we get

⇒ – 70 = – 10x + 50

⇒ 10x = 120

⇒ x = 12

Thus x = – 2, 12

Question 12.

Using determinants, find the area of a triangle whose vertices are (1, 4), (2, 3) and (– 5, – 3). Are the given points collinear?

Answer:

Given: – Vertices are (1, 4), (2, 3) and (– 5, – 3)

Tip: – If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

Expanding along R₁

⇒

Since area can’t be negative

⇒

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero

Now, as the area is not zero

Therefore, Points (1, 4), (2, 3) and (– 5, – 3) are not collinear.

Question 13.

Using determinants, find the area of the triangle with vertices (– 3, 5), (3, – 6) and (7, 2).

Answer:

Given: – Vertices are (– 3, 5), (3, – 6) and (7, 2)

Tip: – If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

Expanding along R₁

⇒

⇒ Δ = 46 sq. units

Question 14.

Using determinants, find the value of k so that the points (k, 2 – 2 k), (– k + 1, 2k) and (– 4 – k, 6 – 2k) may be collinear.

Answer:

Given: – Points are (k, 2 – 2 k), (– k + 1, 2k) and (– 4 – k, 6 – 2k) which are collinear

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero.

If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

Expanding along R₁

⇒

⇒ k(2k – 6 + 2k) – (2 – 2k)(– k + 1 + 4 + k) + 1(6 – 2k – 6k + 2k² + 8k + 2k²) = 0

⇒ 4k² – 6k – 10 + 10k + 6 + 4k² = 0

⇒ 8k² + 4k – 4 = 0

⇒ 8k² + 8k – 4k – 4 = 0

⇒ 8k(k + 1) – 4(k + 1) = 0

⇒ (8k – 4)(k + 1) = 0

If 8k – 4 = 0

⇒

And, If k + 1 = 0

⇒ K = – 1

Hence, k = – 1, 0.5

Question 15.

If the points (x, 2), (5, – 2) and (8, 8) are collinear, find x using determinants.

Answer:

Given: – (x, 2), (5, – 2) and (8, 8) are collinear points

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero.

Now, we know that,

Vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

⇒

Expanding along R₁

⇒

⇒ [x(– 6) + 2(– 3) + 1(24)] = 0

⇒ – 6x – 6 + 24 = 0

⇒ x = 3

Question 16.

If the points (3, – 2), (x,2) and (8,8) are collinear, find x using determinant.

Answer:

Given: – (3, – 2), (x,2) and (8,8) are collinear points

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero

Now, we know that,

Vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

⇒

Expanding along R₁

⇒

⇒ [x(– 6) + 2(x – 8) + 1(8x – 16)] = 0

⇒ – 6x + 2x – 16 + 8x – 16 = 0

⇒ 10x = 50

⇒ x = 5

Question 17.

Using determinants, find the equation of the line joining the points

(1, 2) and (3, 6)

Answer:

Given: – (1, 2) and (3, 6) are collinear points

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero

Now, we know that,

Vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Let, 3^rd point be (x,y)

Substituting given value in above formula

⇒

Expanding along R₁

⇒

⇒ [x(– 4) – y(– 2) + 1(0)] = 0

⇒ – 4x + 2y = 0

⇒ y = 2x

It’s the equation of line

Question 18.

Using determinants, find the equation of the line joining the points

(3, 1) and (9, 3)

Answer:

Given: – (3, 1) and (9, 3) are collinear points

Tip: – For Three points to be collinear, the area of triangle formed by these points will be zero

Now, we know that,

Vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Let, 3^rd point be (x,y)

Substituting given value in above formula

⇒

Expanding along R₁

⇒

⇒ [x(– 2) – y(– 6) + 1(0)] = 0

⇒ – 2x + 6y = 0

⇒ x – 3y = 0

It’s the equation of line

Question 19.

Find values of K, if the area of a triangle is 4 square units whose vertices are

(k,0), (4,0) and (0,2)

Answer:

Given: – Vertices of triangle are (k, 0), (4, 0) and (0, 2) and area of triangle is 4 sq. units

Tip: – If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

⇒

Removing modulus

⇒

Expanding along R₁

⇒

⇒ [k(– 2) – 0(4) + 1(8 – 0)] = ±8

⇒ [ – 2k + 8] = ± 8

Taking + ve sign, we get

⇒ + 8 = – 2x + 8

⇒ – 2k = 0

⇒ k = 0

Taking – ve sign, we get

⇒ – 8 = – 2x + 8

⇒ – 2x = – 16

⇒ x = 8

Thus x = 0, 8

Question 20.

Find values of K, if the area of a triangle is 4 square units whose vertices are

(– 2,0), (0, 4) and (0, k)

Answer:

Given: – Vertices of triangle are (– 2,0), (0, 4) and (0, k) and the area of the triangle is 4 sq. units.

Tip: – If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

⇒

Removing modulus

⇒

Expanding along R₁

⇒

⇒ [ – 2(4 – k) – 0(0) + 1(0 – 0)] = ±8

⇒ – 8 + 2k = ± 8

Taking + ve sign, we get

⇒ 8 = – 8 + 2k

⇒ 2k = 16

⇒ k = 8

Taking – ve sign, we get

⇒ – 8 = 2x – 8

⇒ 2k = 0

⇒ k = 0

Thus k = 0, 8

Exercise 6.4

Question 1.

Solve the following systems of linear equations by Cramer’s rule:

x – 2y = 4

– 3x + 5y = – 7

Answer:

Given: - Two equations x – 2y = 4 and – 3x + 5y = – 7

Tip: - Theorem - Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

x – 2y = 4

– 3x + 5y = – 7

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 5(1) – ( – 3)( – 2)

⇒ D = 5 – 6

⇒ D = – 1

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 5(4) – ( – 7)( – 2)

⇒ D₁ = 20 – 14

⇒ D₁ = 6

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 1( – 7) – ( – 3)(4)

⇒ D₂ = – 7 + 12

⇒ D₂ = 5

Thus by Cramer’s Rule, we have

⇒

⇒ x = – 6

and

⇒

⇒ y = – 5

Question 2.

Solve the following systems of linear equations by Cramer’s rule:

2x – y = 1

7x – 2y = – 7

Answer:

Given: - Two equations 2x – y = 1 and 7x – 2y = – 7

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2x – y = 1

7x – 2y = – 7

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 2( – 2) – (7)( – 1)

⇒ D = – 4 + 7

⇒ D = 3

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 1( – 2) – ( – 7)( – 1)

⇒ D₁ = – 2 – 7

⇒ D₁ = – 9

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 2( – 7) – (7)(1)

⇒ D₂ = – 14 – 7

⇒ D₂ = – 21

Thus by Cramer’s Rule, we have

⇒

⇒ x = – 3

and

⇒

⇒ y = – 7

Question 3.

Solve the following systems of linear equations by Cramer’s rule:

2x – y = 17

3x + 5y = 6

Answer:

Given: - Two equations 2x – y = 17 and 3x + 5y = 6

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2x – y = 17

3x + 5y = 6

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 2(5) – (3)( – 1)

⇒ D = 10 + 3

⇒ D = 13

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 17(5) – (6)( – 1)

⇒ D₁ = 85 + 6

⇒ D₁ = 91

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(6) – (17)(3)

⇒ D₂ = 12 – 51

⇒ D₂ = – 39

Thus by Cramer’s Rule, we have

⇒

⇒ x = 7

and

⇒

⇒ y = – 3

Question 4.

Solve the following systems of linear equations by Cramer’s rule:

3x + y = 19

3x – y = 23

Answer:

Given: - Two equations 3x + y = 19 and 3x – y = 23

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

3x + y = 19

3x – y = 23

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 3( – 1) – (3)(1)

⇒ D = – 3 – 3

⇒ D = – 6

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 19( – 1) – (23)(1)

⇒ D₁ = – 19 – 23

⇒ D₁ = – 42

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(23) – (19)(3)

⇒ D₂ = 69 – 57

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

⇒

⇒ x = 7

and

⇒

⇒ y = – 2

Question 5.

Solve the following systems of linear equations by Cramer’s rule:

2x – y = – 2

3x + 4y = 3

Answer:

Given : – Two equations 2x – y = – 2 and 3x + 4y = 3

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2x – y = – 2

3x + 4y = 3

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 2(4) – (3)( – 1)

⇒ D = 8 + 3

⇒ D = 11

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = – 2(4) – (3)( – 1)

⇒ D₁ = – 8 + 3

⇒ D₁ = – 5

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – ( – 2)(3)

⇒ D₂ = 6 + 6

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

⇒

and

⇒

Question 6.

Solve the following systems of linear equations by Cramer’s rule:

3x + ay = 4

2x + ay = 2, a≠0

Answer:

Given: - Two equations 3x + ay = 4 and 2x + ay = 2, a ≠ 0

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

3x + ay = 4

2x + ay = 2, a≠0

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 3(a) – (2)(a)

⇒ D = 3a – 2a

⇒ D = a

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 4(a) – (2)(a)

⇒ D = 4a – 2a

⇒ D = 2a

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – (2)(4)

⇒ D = 6 – 8

⇒ D = – 2

Thus by Cramer’s Rule, we have

⇒

⇒ x = 2

and

⇒

Question 7.

Solve the following systems of linear equations by Cramer’s rule:

2x + 3y = 10

x + 6y = 4

Answer:

Given: - Two equations 2x – 3y = 10 and x + 6y = 4

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2x + 3y = 10

x + 6y = 4

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 2(6) – (3)(1)

⇒ D = 12 – 3

⇒ D = 9

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10(6) – (3)(4)

⇒ D = 60 – 12

⇒ D = 48

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(4) – (10)(1)

⇒ D₂ = 8 – 10

⇒ D₂ = – 2

Thus by Cramer’s Rule, we have

⇒

and

⇒

Question 8.

Solve the following systems of linear equations by Cramer’s rule:

5x + 7y = – 2

4x + 6y = – 3

Answer:

Given: - Two equations 5x + 7y = – 2 and 4x + 6y = – 3

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

5x + 7y = – 2

4x + 6y = – 3

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 5(6) – (7)(4)

⇒ D = 30 – 28

⇒ D = 2

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = – 2(6) – (7)( – 3)

⇒ D₁ = – 12 + 21

⇒ D₁ = 9

and

Solving determinant, expanding along 1^st row

⇒ D₂ = – 3(5) – ( – 2)(4)

⇒ D₂ = – 15 + 8

⇒ D₂ = – 7

Thus by Cramer’s Rule, we have

⇒

and

⇒

Question 9.

Solve the following systems of linear equations by Cramer’s rule:

9x + 5y = 10

3y – 2x = 8

Answer:

Given: - Two equations 9x + 5y = 10 and 3y – 2x = 8

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

9x + 5y = 10

3y – 2x = 8

So by comparing with the theorem, let's find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 3(9) – (5)( – 2)

⇒ D = 27 + 10

⇒ D = 37

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10(3) – (8)(5)

⇒ D₁ = 30 – 40

⇒ D₁ = – 10

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 9(8) – (10)( – 2)

⇒ D₂ = 72 + 20

⇒ D₂ = 92

Thus by Cramer’s Rule, we have

⇒

and

⇒

Question 10.

Solve the following systems of linear equations by Cramer’s rule:

x + 2y = 1

3x + y = 4

Answer:

Given: - Two equations x + 2y = 1 and 3x + y = 4

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

x + 2y = 1

3x + y = 4

So by comparing with theorem, lets find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 1(1) – (3)(2)

⇒ D = 1 – 6

⇒ D = – 5

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 1(1) – (2)(4)

⇒ D₁ = 1 – 8

⇒ D₁ = – 7

and

Solving determinant, expanding along 1^st row

⇒ D₂ = 1(4) – (1)(3)

⇒ D₂ = 4 – 3

⇒ D₂ = 1

Thus by Cramer’s Rule, we have

⇒

and

⇒

Question 11.

Solve the following system of the linear equations by Cramer’s rule:

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

Answer:

Given: - Equations are: –

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

So by comparing with the theorem, let's find D, D₁, D₂ and D₃

Solving determinant, expanding along 1^st row

⇒ D = 3[( – 4)( – 3) – (3)(1)] – 1[(2)( – 3) – 12] + 1[2 – 4( – 4)]

⇒ D = 3[12 – 3] – [ – 6 – 12] + [2 + 16]

⇒ D = 27 + 18 + 18

⇒ D = 63

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 2[( – 4)( – 3) – (3)(1)] – 1[( – 1)( – 3) – ( – 11)(3)] + 1[( – 1) – ( – 4)( – 11)]

⇒ D₁ = 2[12 – 3] – 1[3 + 33] + 1[ – 1 – 44]

⇒ D₁ = 2[9] – 36 – 45

⇒ D₁ = 18 – 36 – 45

⇒ D₁ = – 63

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 3[3 + 33] – 2[ – 6 – 12] + 1[ – 22 + 4]

⇒ D₂ = 3[36] – 2( – 18) – 18

⇒ D₂ = 126

And,

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 3[44 + 1] – 1[ – 22 + 4] + 2[2 + 16]

⇒ D₃ = 3[45] – 1( – 18) + 2(18)

⇒ D₃ = 135 + 18 + 36

⇒ D₃ = 189

Thus by Cramer’s Rule, we have

⇒

⇒ x = – 1

again,

⇒

⇒ y = 2

and,

⇒

⇒ z = 3

Question 12.

Solve the following system of the linear equations by Cramer’s rule:

x – 4y – z = 11

2x – 5y + 2z = 39

– 3x + 2y + z = 1

Answer:

Given: - Equations are: –

x – 4y – z = 11

2x – 5y + 2z = 39

– 3x + 2y + z = 1

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

x – 4y – z = 11

2x – 5y + 2z = 39

– 3x + 2y + z = 1

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 1[( – 5)(1) – (2)(2)] + 4[(2)(1) + 6] – 1[4 + 5( – 3)]

⇒ D = 1[ – 5 – 4] + 4[8] – [ – 11]

⇒ D = – 9 + 32 + 11

⇒ D = 34

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 11[( – 5)(1) – (2)(2)] + 4[(39)(1) – (2)(1)] – 1[2(39) – ( – 5)(1)]

⇒ D₁ = 11[ – 5 – 4] + 4[39 – 2] – 1[78 + 5]

⇒ D₁ = 11[ – 9] + 4(37) – 83

⇒ D₁ = – 99 – 148 – 45

⇒ D₁ = – 34

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 1[39 – 2] – 11[2 + 6] – 1[2 + 117]

⇒ D₂ = 1[37] – 11(8) – 119

⇒ D₂ = – 170

And,

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 1[ – 5 – (39)(2)] – ( – 4)[2 – (39)( – 3)] + 11[4 – ( – 5)( – 3)]

⇒ D₃ = 1[ – 5 – 78] + 4(2 + 117) + 11(4 – 15)

⇒ D₃ = – 83 + 4(119) + 11( – 11)

⇒ D₃ = 272

Thus by Cramer’s Rule, we have

⇒

⇒ x = – 1

again,

⇒

⇒ y = – 5

and,

⇒

⇒ z = 8

Question 13.

Solve the following system of the linear equations by Cramer’s rule:

6x + y – 3z = 5

X + 3y – 2z = 5

2x + y + 4z = 8

Answer:

Given: - Equations are: –

6x + y – 3z = 5

X + 3y – 2z = 5

2x + y + 4z = 8

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 6[(4)(3) – (1)( – 2)] – 1[(4)(1) + 4] – 3[1 – 3(2)]

⇒ D = 6[12 + 2] – [8] – 3[ – 5]

⇒ D = 84 – 8 + 15

⇒ D = 91

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5[(4)(3) – ( – 2)(1)] – 1[(5)(4) – ( – 2)(8)] – 3[(5) – (3)(8)]

⇒ D₁ = 5[12 + 2] – 1[20 + 16] – 3[5 – 24]

⇒ D₁ = 5[14] – 36 – 3( – 19)

⇒ D₁ = 70 – 36 + 57

⇒ D₁ = 91

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 6[20 + 16] – 5[4 – 2( – 2)] + ( – 3)[8 – 10]

⇒ D₂ = 6[36] – 5(8) + ( – 3)( – 2)

⇒ D₂ = 182

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 6[24 – 5] – 1[8 – 10] + 5[1 – 6]

⇒ D₃ = 6[19] – 1( – 2) + 5( – 5)

⇒ D₃ = 114 + 2 – 25

⇒ D₃ = 91

Thus by Cramer’s Rule, we have

⇒

⇒ x = 1

again,

⇒

⇒ y = 2

and,

⇒

⇒ z = 1

Question 14.

Solve the following system of the linear equations by Cramer’s rule:

x + y = 5

y + z = 3

x + z = 4

Answer:

Given: - Equations are: –

x + y = 5

y + z = 3

x + z = 4

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

x + y = 5

y + z = 3

x + z = 4

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 1[1] – 1[ – 1] + 0[ – 1]

⇒ D = 1 + 1 + 0

⇒ D = 2

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5[1] – 1[(3)(1) – (4)(1)] + 0[0 – (4)(1)]

⇒ D₁ = 5 – 1[3 – 4] + 0[– 4]

⇒ D₁ = 5 – 1[ – 1] + 0

⇒ D₁ = 5 + 1 + 0

⇒ D₁ = 6

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 1[3 – 4] – 5[ – 1] + 0[0 – 3]

⇒ D₂ = 1[ – 1] + 5 + 0

⇒ D₂ = 4

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 1[4 – 0] – 1[0 – 3] + 5[0 – 1]

⇒ D₃ = 1[4] – 1( – 3) + 5( – 1)

⇒ D₃ = 4 + 3 – 5

⇒ D₃ = 2

Thus by Cramer’s Rule, we have

⇒

⇒ x = 3

again,

⇒

⇒ y = 2

and,

⇒

⇒ z = 1

Question 15.

Solve the following system of the linear equations by Cramer’s rule:

2y – 3z = 0

X + 3y = – 4

3x + 4y = 3

Answer:

Given: - Equations are: –

2y – 3z = 0

X + 3y = – 4

3x + 4y = 3

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2y – 3z = 0

x + 3y = – 4

3x + 4y = 3

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 0[0] – 2[(0)(1) – 0] – 3[1(4) – 3(3)]

⇒ D = 0 – 0 – 3[4 – 9]

⇒ D = 0 – 0 + 15

⇒ D = 15

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 0[0] – 2[(0)( – 4) – 0] – 3[4( – 4) – 3(3)]

⇒ D₁ = 0 – 0 – 3[ – 16 – 9]

⇒ D₁ = 0 – 0 – 3( – 25)

⇒ D₁ = 0 – 0 + 75

⇒ D₁ = 75

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant

⇒ D₂ = 0[0] – 0[(0)(1) – 0] – 3[1(3) – 3( – 4)]

⇒ D₂ = 0 – 0 + ( – 3)(3 + 12)

⇒ D₂ = – 45

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 0[9 – ( – 4)4] – 2[(3)(1) – ( – 4)(3)] + 0[1(4) – 3(3)]

⇒ D₃ = 0[25] – 2(3 + 12) + 0(4 – 9)

⇒ D₃ = 0 – 30 + 0

⇒ D₃ = – 30

Thus by Cramer’s Rule, we have

⇒

⇒ x = 5

again,

⇒

⇒ y = – 3

and,

⇒

⇒ z = – 2

Question 16.

Solve the following system of the linear equations by Cramer’s rule:

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Answer:

Given: - Equations are: –

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 5[( – 8)( – 6) – ( – 1)(2)] – 7[( – 6)(6) – 3( – 1)] + 1[2(6) – 3( – 8)]

⇒ D = 5[48 + 2] – 7[ – 36 + 3] + 1[12 + 24]

⇒ D = 250 – 231 + 36

⇒ D = 55

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 11[( – 8)( – 6) – (2)( – 1)] – ( – 7)[(15)( – 6) – ( – 1)(7)] + 1[(15)2 – (7)( – 8)]

⇒ D₁ = 11[48 + 2] + 7[ – 90 + 7] + 1[30 + 56]

⇒ D₁ = 11[50] + 7[ – 83] + 86

⇒ D₁ = 550 – 581 + 86

⇒ D₁ = 55

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 5[(15)( – 6) – (7)( – 1)] – 11[(6)( – 6) – ( – 1)(3)] + 1[(6)7 – (15)(3)]

⇒ D_{2 =} 5[ – 90 + 7] – 11[ – 36 + 3] + 1[42 – 45]

⇒ D₂ = 5[ – 83] – 11( – 33) – 3

⇒ D₂ = – 415 + 363 – 3

⇒ D₂ = – 55

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 5[( – 8)(7) – (15)(2)] – ( – 7)[(6)(7) – (15)(3)] + 11[(6)2 – ( – 8)(3)]

⇒ D₃ = 5[ – 56 – 30] – ( – 7)[42 – 45] + 11[12 + 24]

⇒ D₃ = 5[ – 86] + 7[ – 3] + 11[36]

⇒ D₃ = – 430 – 21 + 396

⇒ D₃ = – 55

Thus by Cramer’s Rule, we have

⇒

⇒ x = 1

again,

⇒

⇒ y = – 1

and,

⇒

⇒ z = – 1

Question 17.

Solve the following system of the linear equations by Cramer’s rule:

2x – 3y – 4z = 29

– 2x + 5y – z = – 15

3x – y + 5z = – 11

Answer:

Given: - Equations are: –

2x – 3y – 4z = 29

– 2x + 5y – z = – 15

3x – y + 5z = – 11

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2x – 3y – 4z = 29

– 2x + 5y – z = – 15

3x – y + 5z = – 11

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 2[(5)(5) – ( – 1)( – 1)] – ( – 3)[( – 2)(5) – 3( – 1)] + ( – 4)[( – 2)( – 1) – 3(5)]

⇒ D = 2[25 – 1] + 3[ – 10 + 3] – 4[2 – 15]

⇒ D = 48 – 21 + 52

⇒ D = 79

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 29[(5)(5) – ( – 1)( – 1)] – ( – 3)[( – 15)(5) – ( – 11)( – 1)] + ( – 4)[( – 15)( – 1) – ( – 11)(5)]

⇒ D₁ = 29[25 – 1] + 3[ – 75 – 11] – 4[15 + 55]

⇒ D₁ = 29[24] + 3[ – 86] – 4(70)

⇒ D₁ = 696 – 258 – 280

⇒ D₁ = 158

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 2[( – 15)(5) – ( – 11)( – 1)] – 29[( – 2)(5) – 3( – 1)] + ( – 4)[( – 11)( – 2) – 3( – 15)]

⇒ D₂ = 2[ – 75 – 11] – 29( – 10 + 3) – 4(22 + 45)

⇒ D₂ = 2[ – 86] – 29( – 7) – 4(67)

⇒ D₂ = – 172 + 203 – 268

⇒ D₂ = – 237

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 2[(5)( – 11) – ( – 15)( – 1)] – ( – 3)[( – 11)( – 2) – ( – 15)(3)] + 29[( – 2)( – 1) – (3)(5)]

⇒ D₃ = 2[ – 55 – 15] + 3(22 + 45) + 29(2 – 15)

⇒ D₃ = 2[ – 70] + 3[67] + 29[ – 13]

⇒ D₃ = – 140 + 201 – 377

⇒ D₃ = – 316

Thus by Cramer’s Rule, we have

⇒

⇒ x = 2

again,

⇒

⇒ y = – 3

and,

⇒

⇒ z = – 4

Question 18.

Solve the following system of the linear equations by Cramer’s rule:

x + y = 1

x + z = – 6

x – y – 2z = 3

Answer:

Given: - Equations are: –

x + y = 1

x + z = – 6

x – y – 2z = 3

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

x + y = 1

x + z = – 6

x – y – 2z = 3

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 1[(0)( – 2) – (1)( – 1)] – 1[( – 2)(1) – 1] + 0[ – 1 – 0]

⇒ D = 1[0 + 1] – 1[ – 3] – 0[ – 2]

⇒ D = 1 + 3 + 0

⇒ D = 4

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 1[(0)( – 2) – (1)( – 1)] – 1[( – 2)( – 6) – 3] + 0[6 – 0]

⇒ D₁ = 1[0 + 1] – 1[12 – 3] + 0[6]

⇒ D₁ = 1[1] – 9 + 0

⇒ D₁ = 1 – 9 + 0

⇒ D₁ = – 8

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 1[( – 6)( – 2) – (1)(3)] – 1[( – 2)(1) – 1] + 0[3 + 6]

⇒ D₂ = 1[12 – 3] – 1( – 2 – 1) + 0(9)

⇒ D₂ = 9 + 3

⇒ D₂ = 12

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 1[(0)(3) – ( – 1)( – 6)] – 1[(3)(1) – 1( – 6)] + 1[ – 1 + 0]

⇒ D₃ = 1[0 – 6] – 1(3 + 6) + 1( – 1)

⇒ D₃ = – 6 – 9 – 1

⇒ D₃ = – 16

Thus by Cramer’s Rule, we have

⇒

⇒ x = – 2

again,

⇒

⇒ y = 3

and,

⇒

⇒ z = – 4

Question 19.

Solve the following system of the linear equations by Cramer’s rule:

x + y + z + 1 = 0

ax + by + cz + d = 0

a²x + b²y + c²z + d² = 0

Answer:

Given: - Equations are: –

x + y + z + 1 = 0

ax + by + cz + d = 0

a²x + b²y + c²z + d² = 0

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

x + y + z + 1 = 0

ax + by + cz + d = 0

a²x + b²y + c²z + d² = 0

So by comparing with theorem, lets find D , D₁ , D₂ and D₃

applying,

⇒

Take (b – a) from c₂ , and (c – a) from c₃ common, we get

⇒

Solving determinant, expanding along 1^st Row

⇒ D = (b – a)(c – a)1[c + a – (b + a)]

⇒ D = (b – a)(c – a)(c + a – b – a)

⇒ D = (b – a)(c – a)(c – b)

⇒ D = (a – b)(b – c)(c – a)

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

applying,

⇒

Take (b – d) from c₂ , and (c – d) from c₃ common, we get

⇒

Solving determinant, expanding along 1^st Row

⇒ D₁ = – (b – d)(c – d)1[c + d – (b + d)]

⇒ D₁ = – (b – d)(c – d)(c + d – b – d)

⇒ D₁ = – (b – d)(c – d)(c – b)

⇒ D₁ = – (d – b)(b – c)(c – d)

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

applying,

⇒

Take (d – a) from c₂ , and (c – a) from c₃ common, we get

⇒

Solving determinant, expanding along 1^st Row

⇒ D₂ = – (d – a)(c – a)1[c + a – (d + a)]

⇒ D₂ = – (d – a)(c – a)(c + a – d – a)

⇒ D₂ = – (d – a)(c – a)(c – d)

⇒ D₂ = – (a – d)(d – c)(c – a)

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

applying,

⇒

Take (b – a) from c₂ , and (d – a) from c₃ common, we get

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = – (b – d)(c – d)1[a + d – (b + a)]

⇒ D₃ = – (b – d)(c – d)(a + d – b – a)

⇒ D₃ = – (b – d)(c – d)(d – b)

⇒ D₃ = – (d – b)(b – d)(c – d)

Thus by Cramer’s Rule, we have

⇒

again,

⇒

and,

⇒

Question 20.

Solve the following system of the linear equations by Cramer’s rule:

x + y + z + w = 2

x – 2y + 2z + 2w = – 6

2x + y – 2z + 2w = – 5

3x – y + 3z – 3w = – 3

Answer:

Given: - Equations are: –

x + y + z + w = 2

x – 2y + 2z + 2w = – 6

2x + y – 2z + 2w = – 5

3x – y + 3z – 3w = – 3

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

x + y + z + w = 2

x – 2y + 2z + 2w = – 6

2x + y – 2z + 2w = – 5

3x – y + 3z – 3w = – 3

So by comparing with theorem, lets find D, D₁, D₂,D₃ and D₄

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

applying,

⇒

⇒ D = 1[ – 6 – 88]

⇒ D = – 94

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₁ = – 1{( – 10)[6( – 1) – 2( – 4)] – ( – 4)[( – 9)6 – ( – 4)3] + 0}

⇒ D₁ = – 1{ – 10[ – 6 + 8] + 4[ – 54 + 12]}

⇒ D₁ = – 1{ – 10[2] + 4[ – 42] }

⇒ D₁ = 188

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₂ = – 1{( – 1)[6( – 9) – 3( – 4)] – ( – 10)[0 – 6( – 4)] + 0[0 + 54]}

⇒ D₂ = – 1{ – 1[ – 54 + 12] + 10(24) + 0}

⇒ D₂ = – 282

Again, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₃ = – 1{( – 1)[ – 3 – ( – 9)2] – ( – 4)[0 – 6( – 9)] + ( – 10)[0 + 6]}

⇒ D₃ = – 1{ – 1[15] + 4(54) – 10(6)}

⇒ D₃ = – 1{ – 15 + 216 – 60}

⇒ D₃ = – 141

And, Solve D₄ formed by replacing 4^th column by B matrices

Here

⇒

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₄ = ( – 3)[( – 9)( – 4) – 0] – 1[9 – ( – 4)( – 9)] + ( – 8)[0 – 16]

⇒ D₄ = – 3[36] – 1(9 – 36) – 8( – 16)

⇒ D₄ = – 108 + 27 + 128

⇒ D₄ = 47

Thus by Cramer’s Rule, we have

⇒

⇒ x = – 2

again,

⇒

⇒ y = 3

again,

⇒

And,

⇒

Question 21.

Solve the following system of the linear equations by Cramer’s rule:

2x – 3z + w = 1

x – y + 2w = 1

– 3y + z + w = 1

x + y + z = 1

Answer:

Given: - Equations are: –

2x – 3z + w = 1

x – y + 2w = 1

– 3y + z + w = 1

x + y + z = 1

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2x – 3z + w = 1

x – y + 2w = 1

– 3y + z + w = 1

x + y + z = 1

So by comparing with theorem, lets find D, D₁, D₂,D₃ and D₄

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

applying,

⇒

expanding along 3^rd row

⇒ D = – 1[ – 3 – ( – 6)4]

⇒ D = – 21

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₁ = ( – 1)[(4)( – 1) – 0(4)] – (3)[( – 3)( – 1) – 0] + 1[ – 12 – 4]

⇒ D₁ = – 1[ – 4 – 0] – 3[3 – 0] – 16

⇒ D₁ = 4 – 9 – 16

⇒ D₁ = – 21

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

⇒ D₂ = – 1{( – 1)[1( – 1) – 1(2)] – ( – 5)[0 – 1(2)] + 1[0 – ( – 1)]}

⇒ D₂ = – 1{ – 1[ – 1 – 2] + 5( – 2) + 1}

⇒ D₂ = 6

Again, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

⇒ D₃ = – 1{( – 2)[0 – (1)2] – ( – 1)[ – 2 – ( – 3)(2)] + 1[ – 2 – 0]}

⇒ D₃ = – 1{ – 2[ – 2] + 1( – 2 + 6) + 1( – 2)}

⇒ D₃ = – 1{4 + 4 – 2}

⇒ D₃ = – 6

And, Solve D₄ formed by replacing 4^th column by B matrices

Here

⇒

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

⇒ D₄ = ( – 1){( – 2)[( – 1)1 – 0] – ( – 5)[ – 2 – 0] + ( – 1)[ – 2 – 3]}

⇒ D₄ = ( – 1){2 – 10 + 5}

⇒ D₄ = 3

Thus by Cramer’s Rule, we have

⇒

⇒ x = 1

again,

⇒

again,

⇒

And,

⇒

Question 22.

Show that each of the following systems of linear equations is inconsistent:

2x – y = 5

4x – 2y = 7

Answer:

Given: - Two equation 2x – y = 5 and 4x – 2y = 7

Tip: - We know that

For a system of 2 simultaneous linear equation with 2 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution.

(iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent.

Now,

We have,

2x – y = 5

4x – 2y = 7

Lets find D

⇒

⇒ D = – 4 + 4

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = – 10 + 7

⇒ D₁ = – 3

And, D₂ by replacing 2^nd column by B

Here

⇒

⇒ D₂ = 14 – 20

⇒ D₂ = – 6

So, here we can see that

D = 0 and D₁ and D₂ are non – zero

Hence the given system of equation is inconsistent.

Question 23.

Show that each of the following systems of linear equations is inconsistent:

3x + y = 5

– 6x – 2y = 9

Answer:

Given: - Two equation 3x + y = 5 and – 6x – 2y = 9

Tip: - We know that

For a system of 2 simultaneous linear equation with 2 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution.

(iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent.

Now,

We have,

3x + y = 5

– 6x – 2y = 9

Lets find D

⇒

⇒ D = – 6 – 6

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = – 10 – 9

⇒ D₁ = – 19

And, D₂ by replacing 2^nd column by B

Here

⇒

⇒ D₂ = 27 + 30

⇒ D₂ = 57

So, here we can see that

D = 0 and D₁ and D₂ are non – zero

Hence the given system of equation is inconsistent.

Question 24.

Show that each of the following systems of linear equations is inconsistent:

3x – y + 2z = 3

2x + y + 3z = 5

x – 2y – z = 1

Answer:

Given: - Three equation

3x – y + 2z = 3

2x + y + 3z = 5

x – 2y – z = 1

Tip: - We know that

For a system of 3 simultaneous linear equation with 3 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solutions.

(iii) If D = 0 and at least one of the determinants D₁, D₂ and D₃ is non – zero, then the system is inconsistent.

Now,

We have,

3x – y + 2z = 3

2x + y + 3z = 5

x – 2y – z = 1

Lets find D

⇒

Expanding along 1^st row

⇒ D = 3[ – 1 – 3( – 2)] – ( – 1)[( – 1)2 – 3] + 2[ – 4 – 1]

⇒ D = 3[5] + 1[ – 5] + 2[ – 5]

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = 3[ – 1 – 3( – 2)] – ( – 1)[( – 1)5 – 3] + 2[ – 10 – 1]

⇒ D_{1 =} 3[5] + [ – 8] + 2[ – 11]

⇒ D_{1 =} 15 – 8 – 22

⇒ D₁ = – 15

⇒ D₁ ≠ 0

So, here we can see that

D = 0 and D₁ is non – zero

Hence the given system of equation is inconsistent.

Hence Proved

Question 25.

Show that each of the following systems of linear equations is inconsistent:

x + y + z = 3

2x – y + z = 2

3x + 6y + 5z = 20.

Answer:

Given: - Three equation

x + y + z = 3

2x – y + z = 2

3x + 6y + 5z = 20.

Tip: - We know that

For a system of 3 simultaneous linear equation with 3 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solution.

(iii) If D = 0 and at least one of the determinants D₁, D₂ and D₃ is non – zero, then the system is inconsistent.

Now,

We have,

x + y + z = 3

2x – y + z = 2

3x + 6y + 5z = 20.

Lets find D

⇒

Expanding along 1^st row

⇒ D = 1[ – 5 – 1(6)] – (1)[(5)2 – 3] + 1[12 + 3]

⇒ D = 1[ – 11] – 1[7] + 1[15]

⇒ D = – 3

So, here we can see that

D ≠ 0

Hence the given system of equation is consistent.

Question 26.

Show that each of the following systems of linear equations has infinite number of solutions and solve:

x – y + z = 3

2x + y – z = 2

– x – 2y + 2z = 1

Answer:

Given: - Three equation

x – y + z = 3

2x + y – z = 2

– x – 2y + 2z = 1

Tip: - We know that

For a system of 3 simultaneous linear equation with 3 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solution.

(iii) If D = 0 and at least one of the determinants D₁, D₂ and D₃ is non – zero, then the system is inconsistent.

Now,

We have,

x – y + z = 3

2x + y – z = 2

– x – 2y + 2z = 1

Lets find D

⇒

Expanding along 1^st row

⇒ D = 1[2 – ( – 1)( – 2)] – ( – 1)[(2)2 – (1)] + 1[ – 4 – ( – 1)]

⇒ D = 1[0] + 1[3] + [ – 3]

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = 3[2 – ( – 1)( – 2)] – ( – 1)[(2)2 – ( – 1)] + 1[ – 4 – 1]

⇒ D_{1 =} 3[2 – 2] + [4 + 1] + 1[ – 5]

⇒ D_{1 =} 0 + 5 – 5

⇒ D₁ = 0

Also, D₂ by replacing 2^nd column by B

Here

⇒

⇒ D₂ = 1[4 – ( – 1)(1)] – (3)[(2)2 – (1)] + 1[2 – ( – 2)]

⇒ D_{2 =} 1[4 + 1] – 3[4 – 1] + 1[4]

⇒ D_{2 =} 5 – 9 + 4

⇒ D₂ = 0

Again, D₃ by replacing 3^rd column by B

Here

⇒

⇒ D₃ = 1[1 – ( – 2)(2)] – ( – 1)[(2)1 – 2( – 1)] + 3[2( – 2) – 1( – 1)]

⇒ D_{3 =} [1 + 4] + [2 + 2] + 3[ – 4 + 1]

⇒ D_{3 =} 5 + 4 – 9

⇒ D₃ = 0

So, here we can see that

D = D₁ = D₂ = D₃ = 0

Thus,

Either the system is consistent with infinitely many solutions or it is inconsistent.

Now, by 1^st two equations, written as

x – y = 3 – z

2x + y = 2 + z

Now by applying Cramer’s rule to solve them,

New D and D₁, D₂

⇒

⇒ D = 1 + 2

⇒ D = 3

Again, D₁ by replacing 1^st column with

⇒

⇒ D₁ = 3 – z – ( – 1)(2 + z)

⇒ D₁ = 5

Again, D₂ by replacing 2^nd column with

⇒

⇒ D₂ = 2 + z – 2 (3 – z)

⇒ D₂ = – 4 + 3z

Hence, using Cramer’s rule

⇒

again,

⇒

Let, z = k

Then

And z = k

By changing value of k you may get infinite solutions

Question 27.

Show that each of the following systems of linear equations has infinite number of solutions and solve:

x + 2y = 5

3x + 6y = 15

Answer:

Given: - Two equation x + 2y = 5 and 3x + 6y = 15

Tip: - We know that

For a system of 2 simultaneous linear equation with 2 unknowns

(iv) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(v) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution.

(vi) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent.

Now,

We have,

x + 2y = 5

3x + 6y = 15

Lets find D

⇒

⇒ D = – 6 – 6

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = 30 – 30

⇒ D₁ = 0

And, D₂ by replacing 2^nd column by B

Here

⇒

⇒ D₂ = 15 – 15

⇒ D₂ = 0

So, here we can see that

D = D₁ = D₂ = 0

Thus,

The system is consistent with infinitely many solutions.

Let

y = k

then,

⇒ x + 2y = 5

⇒ x = 5 – 2k

By changing value of k you may get infinite solutions

Question 28.

Show that each of the following systems of linear equations has infinite number of solutions and solve:

x + y – z = 0

x – 2y + z = 0

3x + 6y – 5z = 0

Answer:

Given: - Three equation

x + y – z = 0

x – 2y + z = 0

3x + 6y – 5z = 0

Tip: - We know that

For a system of 3 simultaneous linear equation with 3 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solution.

(iii) If D = 0 and at least one of the determinants D₁, D₂ and D₃ is non – zero, then the system is inconsistent.

Now,

We have,

x + y – z = 0

x – 2y + z = 0

3x + 6y – 5z = 0

Lets find D

⇒

Expanding along 1^st row

⇒ D = 1[10 – (6)1] – (1)[( – 5)1 – (1)3] + ( – 1)[6 – ( – 2)3]

⇒ D = 1[4] – 1[ – 8] – [12]

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

As one column is zero its determinant is zero

⇒ D₁ = 0

Also, D₂ by replacing 2^nd column by B

Here

⇒

As one column is zero its determinant is zero

⇒ D₂ = 0

Again, D₃ by replacing 3^rd column by B

Here

⇒

As one column is zero its determinant is zero

⇒ D₃ = 0

So, here we can see that

D = D₁ = D₂ = D₃ = 0

Thus,

Either the system is consistent with infinitely many solutions or it is inconsistent.

Now, by 1^st two equations, written as

x + y = z

x – 2y = – z

Now by applying Cramer’s rule to solve them,

New D and D₁, D₂

⇒

⇒ D = – 2 – 1

⇒ D = – 3

Again, D₁ by replacing 1^st column with

⇒

⇒ D₁ = – 2z – 1( – z)

⇒ D₁ = – z

Again, D₂ by replacing 2^nd column with

⇒

⇒ D₂ = – z – z

⇒ D₂ = – 2z

Hence, using Cramer’s rule

⇒

Let, z = k

Then

again,

⇒

And z = k

By changing value of k you may get infinite solutions

Question 29.

Show that each of the following systems of linear equations has infinite number of solutions and solve:

2x + y – 2z = 4

x – 2y + z = – 2

5x – 5y + z = – 2

Answer:

Given: - Three equation

2x + y – 2z = 4

x – 2y + z = – 2

5x – 5y + z = – 2

Tip: - We know that

For a system of 3 simultaneous linear equation with 3 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solution.

(iii) If D = 0 and at least one of the determinants D₁, D₂ and D₃ is non – zero, then the system is inconsistent.

Now,

We have,

2x + y – 2z = 4

x – 2y + z = – 2

5x – 5y + z = – 2

Lets find D

⇒

Expanding along 1^st row

⇒ D = 2[ – 2 – ( – 5)(1)] – (1)[(1)1 – 5(1)] + ( – 2)[ – 5 – 5( – 2)]

⇒ D = 2[3] – 1[ – 4] – 2[5]

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = 4[ – 2 – ( – 5)(1)] – (1)[( – 2)1 – ( – 2)(1)] + ( – 2)[( – 2)( – 5) – ( – 2)( – 2)]

⇒ D_{1 =} 4[ – 2 + 5] – [ – 2 + 2] – 2[6]

⇒ D_{1 =} 12 + 0 – 12

⇒ D₁ = 0

Also, D₂ by replacing 2^nd column by B

Here

⇒

⇒ D₂ = 2[ – 2 – ( – 2)(1)] – (4)[(1)1 – (5)] + ( – 2)[ – 2 – 5( – 2)]

⇒ D_{2 =} 2[ – 2 + 2] – 4[ – 4] + ( – 2)[8]

⇒ D_{2 =} 0 + 16 – 16

⇒ D₂ = 0

Again, D₃ by replacing 3^rd column by B

Here

⇒

⇒ D₃ = 2[4 – ( – 2)( – 5)] – (1)[( – 2)1 – 5( – 2)] + 4[1( – 5) – 5( – 2)]

⇒ D_{3 =} 2[ – 6] – [8] + 4[ – 5 + 10]

⇒ D_{3 =} – 12 – 8 + 20

⇒ D₃ = 0

So, here we can see that

D = D₁ = D₂ = D₃ = 0

Thus,

Either the system is consistent with infinitely many solutions or it is inconsistent.

Now, by 1^st two equations, written as

x – 2y = – 2 – z

5x – 5y = – 2 – z

Now by applying Cramer’s rule to solve them,

New D and D₁, D₂

⇒

⇒ D = – 5 + 10

⇒ D = 5

Again, D₁ by replacing 1^st column with

⇒

⇒ D₁ = 10 + 5z – ( – 2)( – 2 – z)

⇒ D₁ = 6 + 3z

Again, D₂ by replacing 2^nd column with

⇒

⇒ D₂ = – 2 – z – 5 ( – 2 – z)

⇒ D₂ = 8 + 4z

Hence, using Cramer’s rule

⇒

again,

⇒

Let, z = k

Then

And z = k

By changing value of k you may get infinite solutions

Question 30.

Show that each of the following systems of linear equations has infinite number of solutions and solve:

x – y + 3z = 6

x + 3y – 3z = – 4

5x + 3y + 3z = 10

Answer:

Given: - Three equation

x – y + 3z = 6

x + 3y – 3z = – 4

5x + 3y + 3z = 10

Tip: - We know that

For a system of 3 simultaneous linear equation with 3 unknowns

(iv) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(v) If D = 0 and D₁ = D₂ = D₃ = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solution.

(vi) If D = 0 and at least one of the determinants D₁, D₂ and D₃ is non – zero, then the system is inconsistent.

Now,

We have,

x – y + 3z = 6

x + 3y – 3z = – 4

5x + 3y + 3z = 10

Lets find D

⇒

Expanding along 1^st row

⇒ D = 1[9 – ( – 3)(3)] – ( – 1)[(3)1 – 5( – 3)] + 3[3 – 5(3)]

⇒ D = 1[18] + 1[18] + 3[12]

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = 6[9 – ( – 3)(3)] – ( – 1)[( – 4)3 – 10( – 3)] + 3[ – 12 – 30]

⇒ D_{1 =} 6[9 + 9] + [ – 12 + 30] + 3[ – 42]

⇒ D_{1 =} 6[18] + 18 – 3[42]

⇒ D₁ = 0

Also, D₂ by replacing 2^nd column by B

Here

⇒

⇒ D₂ = 1[ – 12 – ( – 3)10] – 6[3 – 5( – 3)] + 3[10 – 5( – 4)]

⇒ D_{2 =} [ – 12 + 30] – 6[3 + 15] + 3[10 + 20]

⇒ D_{2 =} 18 – 6[18] + 3[30]

⇒ D₂ = 0

Again, D₃ by replacing 3^rd column by B

Here

⇒

⇒ D₃ = 1[30 – ( – 4)(3)] – ( – 1)[(10 – 5( – 4)] + 6[3 – 15]

⇒ D_{3 =} 1[30 + 12] + 1[10 + 20] + 6[ – 12]

⇒ D_{3 =} 42 + 30 – 72

⇒ D₃ = 0

So, here we can see that

D = D₁ = D₂ = D₃ = 0

Thus,

Either the system is consistent with infinitely many solutions or it is inconsistent.

Now, by 1^st two equations, written as

x – y = 6 – 3z

x + 3y = – 4 + 3z

Now by applying Cramer’s rule to solve them,

New D and D₁, D₂

⇒

⇒ D = 3 + 1

⇒ D = 4

Again, D₁ by replacing 1^st column with

⇒

⇒ D₁ = 18 – 9z – ( – 1)( – 4 + 3z)

⇒ D₁ = 14 – 5z

Again, D₂ by replacing 2^nd column with

⇒

⇒ D₂ = – 4 + 3z – (6 – 3z)

⇒ D₂ = – 10 + 6z

Hence, using Cramer’s rule

⇒

again,

⇒

Let, z = k

Then

And z = k

By changing value of k you may get infinite solutions

Question 31.

A salesman has the following record of sales during three months for three items A,B and C which have different rates of commission.

Find out the rates of commission on items A,B and C by using determinant method.

Answer:

Given: - Record of sales during three months

Let, rates of commissions on items A,B and C be x, y and z respectively.

Now, we can arrange this model in linear equation system

Thus, we have

90x + 100y + 20z = 800

130x + 50y + 40z = 900

60x + 100y + 30z = 850

Here

⇒

Applying,

⇒

Solving determinant, expanding along 2^nd column

⇒ D = 50[( – 50)( – 170) – ( – 200)( – 60)]

⇒ D = 50[8500 – 12000]

⇒ D = – 175000

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Applying,

⇒

Solving determinant, expanding along 2^nd column

⇒ D₁ = 50[( – 1000)( – 500) – ( – 950)( – 60)]

⇒ D₁ = 50[50000 – 57000]

⇒ D₁ = – 350000

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Applying,

⇒

Solving determinant, expanding along 1^st Row

⇒ D₂ = 20[17500 – 52500]

⇒ D₂ = – 700000

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

Applying,

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 50[161500 – 200000]

⇒ D₃ = – 1925000

Thus by Cramer’s Rule, we have

⇒

⇒ x = 2

again,

⇒

⇒ y = 4

and,

⇒

z = 11

Thus rates of commission of items A, B and C are 2%, 4% and 11% respectively.

Question 32.

An automobile company uses three types of steel S₁, S₂ and S₃ for producing three types of cars C₁, C₂ and C₃. Steel requirements (in tons) for each type of cars are given below:

Using Cramer’s rule, find the number of cars of each type which can be produced using 29, 13 and 16 tonnes of steel of three types respectively.

Answer:

Given: - Steel requirement for each car is given

Let, Number of cars produced by steel type C₁, C₂ and C₃ be x, y and z respectively.

Now, we can arrange this model in linear equation system

Thus, we have

2x + 3y + 4z = 29

x + y + 2z = 13

3x + 2y + z = 16

Here

⇒

Applying,

⇒

Solving determinant, expanding along 3^rd column

⇒ D = 1[30 – 25]

⇒ D = 5

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Applying,

⇒

Solving determinant, expanding along 3^rd column

⇒ D₁ = 1[( – 35)( – 3) – ( – 5)( – 19)]

⇒ D₁ = 1[105 – 95]

⇒ D₁ = 10

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Applying,

⇒

Solving determinant, expanding along 3^rd column

⇒ D₂ = 1[190 – 175]

⇒ D₂ = 15

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

Applying,

⇒

Solving determinant, expanding along 1^st column

⇒ D₃ = – 1[ – 23 – ( – 1)3]

⇒ D₃ = 20

Thus by Cramer’s Rule, we have

⇒

⇒ x = 2

again,

⇒

⇒ y = 3

and,

⇒

⇒ z = 4

Thus Number of cars produced by type C₁, C₂ and C₃ are 2, 3 and 4 respectively.

Exercise 6.5

Question 1.

Solve each of the following systems of homogeneous linear equations:

x + y – 2z = 0

2x + y – 3z = 0

5x + 4y – 9z = 0

Answer:

Given Equations:

x + y – 2z = 0

2x + y – 3z = 0

5x + 4y – 9z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 1(1×(– 9) – 4×(– 3)) – 1(2×(– 9) – 5×(– 3)) – 2(4×2 – 5×1)

= 1(– 9 + 12) – 1(– 18 + 15) – 2(8 – 5)

= 1×3 –1 × (– 3) – 2×3

= 3 + 3 – 6

= 0

Since D = 0, so the system of equation has infinite solution.

Now let z = k

⇒ x + y = 2k

And 2x + y = 3k

Now using the cramer’s rule

x = k

similarly,

y = k

Hence, x = y = z = k.

Question 2.

Solve each of the following systems of homogeneous linear equations:

2x + 3y + 4z = 0

X + y + z = 0

2x + 5y – 2z = 0

Answer:

Given Equations:

2x + 3y + 4z = 0

X + y + z = 0

2x + 5y – 2z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 2(1×(– 2) – 1×5) – 3(1×(– 2) – 2×1) + 4(1×5 – 2×1)

= 2(– 2 – 5) – 3(– 2 – 2) + 4(5 – 2)

= 1×(– 7) – 3 × (– 4) + 4×3

= – 7 + 12 + 12

= 17

Since D ≠ 0, so the system of equation has infinite solution.

Therefore the system of equation has only solution as x = y = z = 0.

Question 3.

Solve each of the following systems of homogeneous linear equations:

3x + y + z = 0

x – 4y3z = 0

2x + 5y – 2z = 0

Answer:

Given Equations:

3x + y + z = 0

x – 4y + 3z = 0

2x + 5y – 2z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 3(– 4×(– 2) – 3×5) – 1(1×(– 2) – 3×2) + 1(1×5 – 2×(– 4))

= 3(8 – 15) – 1(– 2 – 6) + 1(5 + 8)

= 3×(– 7) –1 × (– 8) + 1×13

= – 21 + 8 + 13

= 0

Since D = 0, so the system of equation has infinite solution.

Now let z = k

⇒ 3x + y = – k

And x – 4y = – 3k

Now using the cramer’s rule

similarly,

Question 4.

Find the real values of λ for which the followings system of linear equations has non – trivial solutions. Also, find the non – trivial solutions

2λx – 2y + 3z = 0

X + λy + 2z = 0

2x + λz = 0

Answer:

Given Equations:

2λx – 2y + 3z = 0

x + λy + 2z = 0

2x + λz = 0

For trivial solution D = 0

= 2λ (λ×λ – 0×2) + 2(1×λ – 2×2) + 3(1×0 – 2×λ)

= 2λ (λ² – 0) + 2(λ – 4) + 3(0 – 2λ)

= 2λ³ + 2λ – 8 – 6λ

= 2λ³ + 4λ – 8

Now D = 0

2λ³ – 4λ – 8 = 0

2λ³ – 4λ = 8

λ(λ² – 2) = 4

Hence λ = 2

Now let z = k

⇒ 4x – 2y = – 3k

And x + 2y = – 2k

Now using the cramer’s rule

x = – k

similarly,

Question 5.

If a,b,c are non – zero real numbers and if the system of equations

(a – 1) x = y + z

(b – 1) y = z + x

(c – 1) z = x + y

Has a non – trivial solution, then prove that ab + bc + ca = abc.

Answer:

Given Equations:

(a – 1) x = y + z

(b – 1) y = z + x

(c – 1) z = x + y

Rearranging these equations

(a – 1)x – y – z = 0

– x + (b – 1)y – z = 0

– x – y + (c – 1)z = 0

For trivial solution D = 0

= (a – 1)((b – 1)(c – 1) – (– 1)×(– 1)) + 1(– 1(c – 1) – (– 1)×(– 1)) – 1((– 1)×(– 1) + (b – 1))

= (a – 1)(bc –b –c + 1 – 1) + (1 – c – 1) – 1(1 + b – 1))

= (a – 1)(bc –b –c) –c –b

= abc –ab –ac –bc + b + c – b – c

= abc –ab –ac –bc

Now D = 0

⇒ abc – ab – ac – bc = 0

⇒ abc = ab + bc + ac

Hence proved.

Mcq

Question 1.

Mark the correct alternative in the following:

If A and B are square matrices or order 2, then det (A + B) = 0 is possible only when

A. det (A) = 0 or det (B) = 0

B. det (A) + det (B) = 0

C. det (A) = 0 and det (B) = 0

D. A + B = 0

Answer:

We are given that,

Matrices A and B are square matrices.

Order of matrix A = 2

Order of matrix B = 2

Det (A + B) = 0

We need to find the condition at which det (A + B) = 0.

Let,

Matrix A = [a_ij]

Matrix B = [b_ij]

Since their orders are same, we can express matrices A and B as

A + B = [a_ij + b_ij]

⇒ |A + B| = |a_ij + b_ij| …(i)

Also, we know that

Det (A + B) = 0

That is, |A + B| = 0

From (i),

|a_ij + b_ij| = 0

⇒ [a_ij + b_ij] = 0

Each corresponding element is 0.

⇒ A + B = 0

Thus, det (A + B) = 0 is possible when A + B = 0 .

Question 2.

Mark the correct alternative in the following:

Which of the following is not correct?

A. |A| = |A^T|, where A = [a_ij]_{3 × 3}

B. |kA| = k³ |A|, where A = [a_ij]_{3 × 3}

C. If A is a skew-symmetric matrix of odd order, then |A| = 0

D.

Answer:

We are given that,

A = [a_ij]_3×3

That is, order of matrix A = 3

Example:

Let,

Take determinant of A.

Determinant of 3 × 3 matrices is found as,

So,

⇒ |A| = -11

The transpose of a matrix is a new matrix whose rows are the columns of the original.

So,

Determinant of A^T:

So, we can conclude that,

|A| = |A^T|, where A = [a_ij]_3×3.

Option (B) is correct.

|kA| = k³|A|, where A = [a_ij]_3×3

Example:

Let k = 2.

And,

Take Left Hand Side of the equation:

LHS = |kA|

Multiply 2 by each term of the matrix.

⇒ LHS = 4(4 × 2 – 6 × 4) – 6(2 × 2 – 6 × 6) + 8(2 × 4 – 4 × 6)

⇒ LHS = 4(8 – 24) – 6(4 – 36) + 8(8 – 24)

⇒ LHS = 4(-16) – 6(-32) + 8(-16)

⇒ LHS = -64 + 192 – 128

⇒ LHS = 0

Take Right Hand Side of the equation:

RHS = k³|A|

⇒ RHS = 8 [2(2 × 1 – 3 × 2) – 3(1 × 1 – 3 × 3) + 4(1 × 2 – 2 × 3)]

⇒ RHS = 8 [2(2 – 6) – 3(1 – 9) + 4(2 – 6)]

⇒ RHS = 8 [2(-4) – 3(-8) + 4(-4)]

⇒ RHS = 8 [-8 + 24 – 16]

⇒ RHS = 8 × 0

⇒ RHS = 0

Since, LHS = RHS.

We can conclude that,

|kA| = k³|A|, where A = [a_ij]_3×3

Option (C) is also correct.

If A is a skew-symmetric matrix of odd order, then |A| = 0.

If the transpose of a matrix is equal to the negative of itself, the matrix is said to be skew symmetric. In other words, A^T = -A.

Example,

Let a matrix of odd order 3 × 3 be,

Take determinant of A.

Thus, we can conclude that

If A is a skew-symmetric matrix of odd order, then |A| = 0.

Option (D) is incorrect.

Let a = 1, b = 3, c = 3, d = -4, e = -2, f = 5, g = 0 and h = 2.

Take Left Hand Side,

⇒ LHS = 4 × 2 – (-1) × 3

⇒ LHS = 8 + 3

⇒ LHS = 11

Take Right Hand Side,

⇒ RHS = (1 × 0 – 3 × (-2)) + (3 × 2 – (-4) × 5)

⇒ RHS = (0 + 6) + (6 + 20)

⇒ RHS = 6 + 26

⇒ RHS = 32

Since, LHS ≠ RHS. Then, we can conclude that,

Question 3.

Mark the correct alternative in the following:

If and C_ij is cofactor of a_ij in A, then value of |A| is given by

A. a₁₁C₃₁ + a₁₂C₃₂ + a₁₃C₃₃

B. a₁₁C₁₁ + a₁₂C₂₁ + a₁₃C₃₁

C. a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃

D. a₁₁C₁₁ + a₂₁C₂₁ + a₃₁C₃₁

Answer:

Let us understand what cofactor of an element is.

A cofactor is the number you get when you remove the column and row of a designated element in a matrix, which is just a numerical grid in the form of a rectangle or a square. The cofactor is always preceded by a positive (+) or negative (-) sign, depending whether the element is in a + or - position. It is

Let us recall how to find the cofactor of any element:

If we are given with,

Cofactor of any element say a₁₁ is found by eliminating first row and first column.

⇒ Cofactor of a₁₁ = a₂₂ × a₃₃ – a₂₃ × a₃₂

The sign of cofactor of a₁₁ is (+).

And, cofactor of any element, say a₁₂ is found by eliminating first row and second column.

⇒ Cofactor of a₁₂ = a₂₁ × a₃₃ – a₂₃ × a₃₁

The sign of cofactor of a₁₂ is (-).

We are given that,

And C_ij is the cofactor of a_ij in A.

Determinant of 3 × 3 matrix is given as,

Or,

Or using the definition of cofactors,

Thus, proved.

Question 4.

Mark the correct alternative in the following:

Which of the following is not correct in a given determinant of A, where A = [a_ij]_{3 × 3}.

A. Order of minor is less than order of the det (A)

B. Minor of an element can never be equal to cofactor of the same element

C. Value of a determinant is obtained by multiplying elements of a row or column by corresponding cofactors

D. Order of minors and cofactors of elements of A is same

Answer:

For option (A),

A minor is the determinant of the square matrix formed by deleting one row and one column from some larger square matrix.

So, the order of minor is always less than the order of determinant.

Thus, option (A) is correct.

For option (B),

A cofactor is the number you get when you remove the column and row of a designated element in a matrix, which is just a numerical grid in the form of a rectangle or a square.

A minor is the determinant of the square matrix formed by deleting one row and one column from some larger square matrix.

Since, the definition of cofactor and minor is same, then we can conclude that

Minor of an element is always equal to cofactor of the same element.

Thus, option (B) is incorrect.

For option (C),

Determinant of 3 × 3 matrix is given as,

Or,

Or using the definition of cofactors,

Thus, option (C) is correct.

For option (D),

A cofactor is the number you get when you remove the column and row of a designated element in a matrix, which is just a numerical grid in the form of a rectangle or a square.

A minor is the determinant of the square matrix formed by deleting one row and one column from some larger square matrix.

Since, the definition of cofactor and minor is same, then we can say that,

Minor of an element is always equal to cofactor of the same element.

⇒ The order of the minor and cofactor of A is same. (where A is some matrix)

Thus, option (D) is correct.

Question 5.

Mark the correct alternative in the following:

Let . Then, the value of 5a + 4b + 3c + 2d + e is equal to

A. 0

B. -16

C. 16

D. none of these

Answer:

We are given that,

We need to find the value of 5a + 4b + 3c + 2d + e.

Determinant of 3 × 3 matrix is given as,

So,

Since,

⇒ x⁴ – x³ – 12x² + 12x = ax⁴ + bx³ + cx² + dx + e

Comparing the left hand side and right hand side of the equation, we get

a = 1

b = -1

c = -12

d = 12

e = 0

Putting these values in 5a + 4b + 3c + 2d + e, we get

5a + 4b + 3c + 2d + e = 5(1) + 4(-1) + 3(-12) + 2(12) + 0

⇒ 5a + 4b + 3c + 2d + e = 5 – 4 – 36 + 24

⇒ 5a + 4b + 3c + 2d + e = 25 – 36

⇒ 5a + 4b + 3c + 2d + e = -11

Thus, the values of 5a + 4b + 3c + 2d + e is -11.

Question 6.

Mark the correct alternative in the following:

The value of the determinant is independent of

A. n

B. a

C. x

D. none of these

Answer:

Let us solve the determinant.

We know that,

Determinant of 3 × 3 matrix is given as,

So,

By trigonometric identity, we have

sin (α – β) = sin α cos β – cos α sin β

So, we can write

Note that, the result has ‘a’ as well as ‘x’, but doesn’t contain ‘n’.

Thus, the determinant is independent of n.

Question 7.

Mark the correct alternative in the following:

If , then

A. Δ₁ + Δ₂ = 0

B. Δ₁ + 2Δ₂ = 0

C. Δ₁ = Δ₂

D. none of these

Answer:

We are given that,

Let us find the determinants ∆₁ and ∆₂.

We know that,

Determinant of 3 × 3 matrix is given as,

So,

⇒ ∆₁ = (b × c² – c × b²) – (a × c² – c × a²) + (a × b² – b × a²)

⇒ ∆₁ = bc² – b²c – ac² + a²c + ab² – a²b …(i)

Also,

⇒ ∆₂ = (ca × c – b × ab) – bc(1 × c – b × 1) + a(1 × ab – ca × 1)

⇒ ∆₂ = ac² – ab² – bc(c – b) + a(ab – ac)

⇒ ∆₂ = ac² – ab² – bc² + b²c + a²b – a²c …(ii)

Checking Option (A).

Adding ∆₁ and ∆₂ by using values from (i) and (ii),

∆₁ + ∆₂ = (bc² – b²c – ac² + a²c + ab² – a²b) + (ac² – ab² – bc² + b²c + a²b – a²c)

⇒ ∆₁ + ∆₂ = bc² – bc² – b²c + b²c – ac² + ac² + ab² – ab² – a²b + a²b

⇒ ∆₁ + ∆₂ = 0

Thus, option (A) is correct.

Checking Option (B).

Multiplying 2 by (ii),

2∆₂ = 2(ac² – ab² – bc² + b²c + a²b – a²c)

⇒ 2∆₂ = 2ac² – 2ab² – 2bc² + 2b²c + 2a²b – 2a²c …(iii)

Then, adding 2∆₂ with ∆₁,

∆₁ + 2∆₂ = (bc² – b²c – ac² + a²c + ab² – a²b) + (2ac² – 2ab² – 2bc² + 2b²c + 2a²b – 2a²c)

⇒ ∆₁ + 2∆₂ = bc² – 2bc² – b²c + 2b²c – ac² + 2ac² + ab² – 2ab² – a²b + 2a²b

⇒ ∆₁ + 2∆₂ = -bc² + b²c + ac² – ab² + a²b

⇒ ∆₁ + 2∆₂ ≠ 0

Thus, option (B) is not correct.

Checking option (C).

Obviously, ∆₁ ≠ ∆₂

Since, by (i) and (ii), we can notice ∆₁ and ∆₂ have different values.

Thus, option (C) is not correct.

Question 8.

Mark the correct alternative in the following:

If and , then n equals

A. 4

B. 6

C. 8

D. none of these

Answer:

We are given that,

We need to find the value of e.

We know that,

Determinant of 3 × 3 matrix is given as,

Question 9.

Mark the correct alternative in the following:

Let be an identify in x, were a, b, c, d, e are independent of x. Then the value of e is

A. 4

B. 0

C. 1

D. none of these

Answer:

We are given that,

We need to find the value of e.

We know that,

Determinant of 3 × 3 matrix is given as,

So,

⇒ (x² + 3x)[-2x × 3x – (x – 4)(x + 4)] – (x – 1)[(x + 1) × 3x – (x – 4)(x – 3)] + (x + 3)[(x + 1)(x + 4) – (-2x)(x – 3)] = ax⁴ + bx³ + cx² + dx + e

⇒ (x² + 3x)[-6x – (x² – 16)] – (x – 1)[3x(x + 1) – (x² – 3x – 4x + 12)] + (x + 3)[x² + x + 4x + 4 + 2x(x – 3)] = ax⁴ + bx³ + cx² + dx + e

⇒ (x² + 3x)[-6x – x² + 16] – (x – 1)[3x² + 3x – x² + 7x – 12] + (x + 3)[x² + 5x + 4 + 2x² – 6x] = ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 6x³ + 16x² – 3x³ – 18x² + 48x – (x – 1)[2x² + 10x – 12] + (x + 3)[3x² – x + 4] = ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 9x³ – 2x² + 48x – (2x³ – 2x² + 10x² – 10x – 12x + 12) + 3x³ + 9x² – x² – 3x + 4x + 12 = ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 9x³ – 2x² + 48x – 2x³ + 2x² – 10x² + 10x + 12x – 12 + 3x³ + 9x² – x² – 3x + 4x + 12 = ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 9x³ – 2x³ + 3x³ – 2x² + 2x² + 9x² – x² + 48x + 10x + 12x – 3x + 4x – 12 + 12 = ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 8x³ + 8x² + 23x + 0 = ax⁴ + bx³ + cx² + dx + e

Comparing left hand side and right-hand side of the equation, we get

e = 0

Thus, e = 0.

Question 10.

Mark the correct alternative in the following:

Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant . The other factor in the value of the determinant is

A. 4

B. 2

C. a + b + c

D. none of these

Answer:

Let assume a=0, b=1, c=2

Now expending around colume1

0-1(-4-6)+2(3+4)=k(1)(3)(2)

6k=24

K=4

Question 11.

Mark the correct alternative in the following:

If a, b, c are distinct then the value of x satisfying is

A. c

B. a

C. b

D. 0

Answer:

(this is possible when x=0)

Question 12.

Mark the correct alternative in the following:

If the determinant , then

A. a, b, c are in H.P.

B. α is a root of 4ax² + 12bx + 9c = 0 or, a, b, c are in G.P.

C. a, b, c are in G.P. only

D. a, b, c are in A.P.

Answer:

expend the determinats

a[-(2bα+3c)² ]-b[-(2bα+3c)(2aα+3b)]+ (2aα+3b)[b(2bα+3c)-c(2aα+3b)]=0

-a(2bα+3c)² + b(2bα+3c)(2aα+3b)+(2aα+3b)[2b^2 α+3bc-3bc-2acα]=0

(2bα+3c) [-2abα-3ac+2abα+3b² ]+ (2aα+3b)(2α)( b² -ac)=0

(2bα+3c) [-3ac +3b² ]+ (2aα+3b)(2α)( b²-ac)=0

(b² – ac)[4aα² + 12bα + ac] = 0=

CASE1→(b² -ac)=0

b² =ac {abc are in Gp}

CASE2→(4aα² +12bα+ac)=0 {Whose one root is }

Question 13.

Mark the correct alternative in the following:

If ω is a non-real cube root of unity and n is not a multiple of 3, then is equal to

A. 0

B. ω

C. ω²

D. 1

Answer:

Assume that n=2(not multiple of 3)

expend the determinant

∆=1(1-w³)-w² (w-w⁴ )+w(w² -w² )

∆=1- w³- w³+w⁶ + w³- w³

∆=0

Question 14.

Mark the correct alternative in the following:

If , then the value of is

A. n

B. 2n

C. -2n

D. n²

Answer:

assume (n)=1

1(4-1)-1(8-1) +2(2-1)=-2

Answer=c(-2n)

Question 15.

Mark the correct alternative in the following:

If a > 0 and discriminant of ax² + 2bx + c is negative, then is

A. positive

B. (ac – b²)(ax² + 2bx + c)

C. negative

D. 0

Answer:

discriminant of ax² + 2bx + c =0

and ax² + 2bx + c

R₃→ R₃ – X R₁ – R₂

- (2ax + 2bx +c)(-b² + ac) < 0

Question 16.

Mark the correct alternative in the following:

The value of is

A. 5²

B. 0

C. 5¹³

D. 5⁹

Answer:

∆=0

Question 17.

Mark the correct alternative in the following:

A. 7

B. 10

C. 13

D. 17

Answer:

=10

Question 18.

Mark the correct alternative in the following:

If a, b, c are in A.P., then the determinant

A. 0

B. 1

C. x

D. 2x

Answer:

{a + c=2b}

R₁→ R₁ – R₂

R₂→ R₂ – R₃

R₁→ R₁ – R₂

∆=0

Question 19.

Mark the correct alternative in the following:

If A + B + C = π, then the value of is equal to

A. 0

B. 1

C. 2sin B tan A cos C

D. none of these

Answer:

∆=0

Question 20.

Mark the correct alternative in the following:

The number of distinct real roots of lies in the interval is

A. 1

B. 2

C. 3

D. 0

Answer:

∆=(csc x+2 sec x )[(csc x-sec x)² ]

Case1: (csc x+2 sec x )=0

(1^st real root)

Case:(csc x-sec x)² =0

Tan x=1 (2^nd real root)

Question 21.

Mark the correct alternative in the following:

Let , where 0 ≤ θ ≤ 2π. Then,

A. Det (A) = 0

B. Det (A) ∈ (2, ∞)

C. Det (A) ∈ (2, 4)

D. Det (A) ∈ [2, 4]

Answer:

A=2[(+1] 0≤(sin θ)² ≤1

A∈2[1,2]

A∈[2,4]

Question 22.

Mark the correct alternative in the following:

If , then x =

A. 3

B. ± 3

C. ± 6

D. 6

Answer:

2x² -40=18+14

x=±6

Question 23.

Mark the correct alternative in the following:

If , then

A. f(a) = 0

B. 3bc

C. a³ + b³ + c³ – 3abc

D. none of these

Answer:

ON TRANSPOSING

Question 24.

Mark the correct alternative in the following:

The value of the determinant is

A. a³ + b³ + c³

B. 3bc

C. a³ + b³ + c³ – 3abc

D. none of these

Answer:

assume a=1,b=2, c=3 (put in determinant)

∆=[-1(12-6)-5(3-4)+1(3-6)]

∆=-4

put a=1,b=2, c=3 in option A,B,C,D

ANSWER=D(none of these )

Question 25.

Mark the correct alternative in the following:

If x, y, z are different from zero and , then the value of is

A. xyz

B.

C. –x –y – z

D. –1

Answer:

x[(1+y)z+ y]-1[-yz]=0

xz + xyz+ xy+yz =0 (divide by xyz in both side)

Question 26.

Mark the correct alternative in the following:

The determinant equals

A. abc(b – c)(c – a)(a – b)

B. (b – c)(c – a)(a – b)

C. (a + b + c)(b – c)(c – a)(a – b)

D. none of these

Answer:

assume a=1,b=2, c=3 (put in determinant)

puta=1,b=2, c=3 in option A,B,C,D

ANSWER=D(none of these )

Question 27.

Mark the correct alternative in the following:

If x, y ∈ R, then the determinant lies in the interval

A. B. [–1, 1]

C. D.

Answer:

Δ = ( sin y – cos y)[(cos x )² + ( sin x)²]

= ( sin y – cos y)

= - ( cos y – sin y)

Question 28.

Mark the correct alternative in the following:

The maximum value of is (θ is real)

A.

B.

C.

D.

Answer:

∆=cos θ (1-1-sin θ)

∆=-cos θ sin θ

-1 ≤sin 2θ≤1

[]

Question 29.

Mark the correct alternative in the following:

The value of the determinant is

A. 9x²(x + y)

B. 9y²(x + y)

C. 3y²(x + y)

D. 7x²(x + y)

Answer:

Question 30.

Mark the correct alternative in the following:

Let , then is equal to

A. 0

B. –1

C. 2

D. 3

Answer:

ANSWER= -1+1=0

Question 31.

Mark the correct alternative in the following:

There are two values of a which makes the determinant equal to 86. The sum of these two values is

A. 4

B. 5

C. –4

D. 9

Answer:

Sum of roots = [b=1and a=1]

Sum of roots= -4

Question 32.

Mark the correct alternative in the following:

If , t hen the value of is

A. 4

B. 8

C. 16

D. 32

Answer:

2×16=32

Question 33.

Mark the correct alternative in the following:

The value of is

A. 2

B. 4

C. 8

D. n²

Answer:

∆= 1/2 [8n+20-8n-4]

∆=8

Very Short Answer

Question 1.

If A is a singular matrix, then write the value of |A|.

Answer:

Since a singular matrix is a matrix whose determinant is 0, Therefore the determinant of A is 0.

Question 2.

Write the value of the determinant .

Answer:

Let

Using the property that if the equimultiples of corresponding elements of other rows (or columns) are added to every element of any row (or column) of a determinant, then the value of determinant remains the same

Using row transformation, R₂→R₂-xR₁

Using the property that if all elements of any row or column of a determinant are 0, then the value of determinant is 0.

Since R₂ has all elements 0, therefore ∆=0.

Question 3.

State whether the matrix is singular or non-singular.

Answer:

Let

Then

=2×4-3×6

=-10 (Expanding along R₁)

Since |A|≠0, therefore A is a non-singular matrix.

Question 4.

Find the value of the determinant .

Answer:

Let

Using the property that if some or all elements of a row or column of a determinant are expressed as the sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants.

We get,

Using the property that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Hence,

=0×3-1×2=-2 (Expanding along R₁)

Question 5.

Find the value of the determinant .

Answer:

Let

We get,

Using the property that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

We get,

Using row transformation, R₂→R₂-R₁ and R₃=R₃-R₁

We get,

Using the property that if each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

Taking out factor 2 from R₃,

We get,

Using the property that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Since, R₂ and R₃ are identical, therefore ∆ =0.

Question 6.

Write the value of the determinant .

Answer:

Let

Using column transformation, C₁→C₁+C₃

We get,

Using the property that if each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

Taking out factor(a+b+c) from C₁,

We get,

Using column transformation, C₁→C₁-C₂

We get,

Expanding along C₁, we get

∆ =(a + b + c)×[(1-a)(c + a-(b + c))]=(1-a)(a-b)(a + b + c)

Question 7.

If and , find the value of |A| + |B|.

Answer:

Given that and , we have to find |A|+|B|

Then, and

|A|=0×1-i×i

=-i²

=1 (Expanding along R₁ and since i²=-1)

|B|=0×1-1×1

=-1 (Expanding along R₁)

|A|+|B|=1-1

Question 8.

If and , find |AB|.

Answer:

Given that and , we have to find |AB|

Then, and

|A|=1×-1-2×3

=-7 (Expanding along R₁)

|B|=1×0-0×-1

=0 (Expanding along R₁)

Since |AB|=|A||B|,

Therefore |AB|=-7×0=0

Question 9.

Evaluate: .

Answer:

Let

We get,

Using the property that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Hence,

=0×6-2×4=-8 (Expanding along R₁)

Question 10.

If w is an imaginary cube root of unity, find the value of .

Answer:

Let

Using row transformation, R₂→R₂-ωR₁

(Since, ω is a cube root of 1, therefore ω³=1)

Using the property that if all elements of a row or column of a determinant are 0, the value of determinant is 0.

Hence ∆=0

Question 11.

If and , find .

Answer:

Given that and , we have to find |AB|

Then, and

|A|=1×-1-2×3

=-7 (Expanding along R₁)

|B|=1×(-2)-(-4)×3

=10 (Expanding along R₁)

Since |AB|=|A||B|,

Therefore |AB|=-7×10=-70

Question 12.

If A = [a_ij] is a 3 × 3 diagonal matrix such that a₁₁ = 1, a₂₂ = 2 and a₃₃ = 3, then find |A|.

Answer:

Since A is a diagonal matrix, therefore, all it’s non-diagonal members are 0. And a₁₁=1, a₂₂=2 and a₃₃=3

We get

Then,

Expanding along R₁

|A|= 1(2×3-0)=6

Question 13.

If A = [a_ij] is a 3 × 3 scalar matrix such that a₁₁ = 2, then write the value of |A|.

Answer:

A scalar matrix is a matrix of order m which is equal to a constant λ multiplied with the Identity matrix of order m.

Since a₁₁=2, hence λ=2 and m=3

Hence

Then,

Expanding along R₁

|A|=2(2×2-0)

Question 14.

If I₃ denotes identity matrix of order 3 × 3, write the value of its determinant.

Answer:

Then,

Expanding along R₁

|I₃|=1(1×1-0)

Question 15.

A matrix A of order 3 × 3 has determinant 5. What is the value of |3A|?

Answer:

If the determinant of a matrix A of order m is ∆, then the determinant of matrix λA, where λ is a scalar, is λ^m∆.

In this question, ∆=5, λ=3 and m=3.

|λA|=3³×5

=135

Question 16.

On expanding by first row, the value of the determinant of 3 × 3 square matrix A = [a_ij] is a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃, where C_ij is the cofactor of a_ij is the cofactor of a_ij in A. Write the expression for its value of expanding by second column.

Answer:

The value of determinant written in the form of cofactors is equal to the sum of products of elements of that row (or column) multiplied by their corresponding cofactors.

Hence, the value of determinant |A|, of matrix A=[a_ij] of order 3×3, expanded along column 2 will be

|A|=a₁₂×C₁₂+a₂₂×C₂₂+a₃₂×C₃₂

Question 17.

Let A = [a_ij] be a square matrix of order 3 × 3 and C_ij denote cofactor of a_ij in A. If |A| = 5, write the value of a₃₁C₃₁ + a₃₂C₃₂ + a₃₃C₃₃.

Answer:

The value of determinant |A||, of matrix A=[a_ij] of order 3×3, is given to be 5.

The value of determinant written in the form of cofactors is equal to the sum of products of elements of that row (or column) multiplied by their corresponding cofactors.

The value of |A| expanded along row 3 will be

|A|= a₃₁×C₃₁+a₃₂×C₃₂+a₃₃×C₃₃, which is the required expression

Hence, the value of required expression is equal to |A|=5.

Question 18.

In question 18, write the value of a₁₁C₂₁ + a₁₂C₂₂ + a₁₃C₂₃.

Answer:

We have to find out the value of a₁₁×C₂₁+a₁₂×C₂₂+a₁₃×C₂₃

LetI=a₁₁×C₂₁+a₁₂×C₂₂+a₁₃×C₂₃

I= a₁₁×-(a₁₂a₃₃-a₁₃a₃₂)+a₁₂×(a₁₁a₃₃-a₁₃a₃₁)+a₁₃×-(a₁₁a₃₂-a₁₂a₃₁)

I=-a₁₁a₁₂a₃₃+a₁₁a₁₃a₃₂+a₁₁a₁₂a₃₃-a₁₂a₁₃a₃₁-a₁₁a₁₃a₃₂+a₁₂a₁₃a₃₁

I= a₁₁a₁₂a₃₃–a₁₁a₁₂a₃₃+a₁₁a₁₃a₃₂-a₁₁a₁₃a₃₂+ a₁₂a₁₃a₃₁- a₁₂a₁₃a₃₁

I=0

Question 19.

Write the value of .

Answer:

Let

Expanding along R₁,

we get ∆= sin20°cos70°-(-cos20°)sin70°

= sin20°cos70°+cos20°sin70°

Since sin(A+B)= sinAcosB+cosAsinB

Hence, sin20°cos70°+cos20°sin70°=sin(20°+70°)

=sin(90°)

Hence, ∆=1

Question 20.

If A is a square matrix satisfying A^TA = I, write the value of |A|.

Answer:

Since A^TA=I

Taking determinant both sides

|A^TA|=|I|

Using |AB|=|A||B|,

|A^T|=|A| and |I|=1, we get

|A||A|=1

(|A|)²=1

Hence, |A|=±1

Question 21.

If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.

Answer:

Given that |A|=3 and AB=I

Since AB=I

Taking determinant both sides

|AB|=|I|

Using |AB|=|A||B|,|A|=3 and |I|=1, we get

3|B|=1

Hence,

Question 22.

A is skew-symmetric of order 3, write the value of |A|.

Answer:

Since A is a skew-symmetric matrix, Therefore

A^T=-A

Taking determinant both sides

|A^T|=|-A|

Using |A^T|=|A| and |λA|=λ^m|A| where m is the order of A

|A|=(-1)³|A|

=-|A| or 2|A|=0

Hence, |A|=0

Question 23.

If A is a square matrix of order 3 with determinant 4, then write the value of |-A|.

Answer:

Since |λA|= λ^m|A|

Given that λ=-1, m=3 and |A|=4, we get

|-A|=(-1)³×4=-4

Question 24.

If A is a square matrix such that |A| = 2, write the value of |AA^T|.

Answer:

Given that |A|=2, we have to find |AA^T|

Using |AB|=|A||B| and |A^T|=|A|, we get

|AA^T|=|A||A^T|

=|A||A|

=2×2

Question 25.

Find the value of the determinant .

Answer:

Let

Using row transformation, R₁→R₁-3R₂

We get,

Using the property that if all elements of a row or column of a determinant are 0, the value of determinant is 0.

Hence ∆=0

Question 26.

Write the value of the determinant .

Answer:

Let

Using the property that if each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

Taking out factor 2 from R₂ and 3 from R₃,

We get,

Using the property that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Since R₁, R₂ and R₃ are identical, therefore ∆=0

Question 27.

If the matrix is singular, find the value of x.

Answer:

Let

Then,

=5x×1-2×-10 (Expanding along R₁)

|A|=5x+20

For A to be singular, |A|=0

Hence 5x+20=0 or x=-4

Question 28.

If A is a square matrix of order n × n such that |A| = λ, then write the value of |-A|.

Answer:

Since |kA|= k^m|A|

Given that k=-1, m=n and |A|=λ, we get

|-A|=(-1)ⁿ×λ

Hence, |-A|=λ if n is even and |-A|=-λ if n is odd.

Question 29.

Find the value of the determinant .

Answer:

Let

Using the property that if each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

Taking out factor 2² from R₁ and 2³ from R₂,

Using the property that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Since R₁ and R₂ are identical, therefore ∆=0.

Question 30.

If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.

Answer:

We are given that,

A = non-singular matrix

B = non-singular matrix

Order of A = Order of B

We need to find whether AB is singular or non-singular.

Let us recall the definition of non-singular matrix.

Non-singular matrix, also called regular matrix, is a square matrix that is not singular, i.e., one that has a matrix inverse.

We can say that, a square matrix A is non-singular matrix iff its determinant is non-zero, i.e., |A| ≠ 0.

While a singular matrix is a square matrix that doesn’t have a matrix inverse. Also, the determinant is zero, i.e., |A| = 0.

So,

By definition, |A| ≠ 0 and |B| ≠ 0 since A and B are non-singular matrices.

Let,

Order of A = Order of B = n × n

⇒ Matrices A and B can be multiplied

⇒ A × B = AB

If we have matrices A and B of same order then we can say that,

|AB| = 0 iff either |A| or |B| = 0.

And it is clear that, |A|, |B| ≠ 0.

⇒ |AB| ≠ 0

And if |AB| ≠ 0, then by definition AB is s non-singular matrix.

Thus, AB is a singular matrix.

Question 31.

A matrix of order 3 × 3 has determinant 2. What is the value of |A(3I)|, where I is the identity matrix of order 3 × 3.

Answer:

We are given that,

Order of a matrix = 3 × 3

Determinant = 2

I = Identity matrix of order 3 × 3

We need to find the value of |A(3I)|.

Let the given matrix be A.

Then, |A| = 2

Also, since I is an identity matrix, then

⇒ Det (I) = 1

Or,

|I| = 1

Then, we can say

3(I) = 3

⇒ 3I = 3

Thus,

|A(3I)| = |A(3)| [∵, 3I = 3]

⇒ |A(3I)| = |3A|

By property of determinants, we know that

|KA| = Kⁿ|A|, if A is of n^th order.

⇒ |A(3I)| = 3³|A| [∵, A has an order of 3 × 3 ⇒ |3A| = 3³ |A|]

⇒ |A(3I)| = 27 |A|

Since, |A| = 2. Then,

⇒ |A(3I)| = 27 × 2

⇒ |A(3I)| = 54

Thus, |A(3I)| = 54.

Question 32.

If A and B are square matrices of order 3 such that |A| = -1, |B| = 3, then find the value of |3AB|.

Answer:

We are given that,

A and B are square matrices of order 3.

|A| = -1, |B| = 3

We need to find the value of |3AB|.

By property of determinant,

|KA| = Kⁿ|A|

If A is of n^th order.

If order of A = 3 × 3

And order of B = 3 × 3

⇒ Order of AB = 3 × 3 [∵, Number of columns in A = Number of rows in B]

We can write,

|3AB| = 3³|AB| [∵, Order of AB = 3 × 3]

Now, |AB| = |A||B|.

⇒ |3AB| = 27|A||B|

Putting |A| = -1 and |B| = 3, we get

⇒ |3AB| = 27 × -1 × 3

⇒ |3AB| = -81

Thus, the value of |3AB| = -81.

Question 33.

Write the value of .

Answer:

We need to find the value of

Determinant of 2 × 2 matrix is found as,

So,

Rearranging,

Using the algebraic identity,

(x + y)(x – y) = x² – y²

Here, i is iota, an imaginary number.

Note that,

i² = -1

So,

Thus,

Question 34.

Write the cofactor of a₁₂ in the matrix .

Answer:

We need to find the cofactor of a₁₂ in the matrix

Let us recall how to find the cofactor of any element:

If we are given with,

Cofactor of any element, say a₁₁ is found by eliminating first row and first column.

⇒ Cofactor of a₁₁ = a₂₂ × a₃₃ – a₂₃ × a₃₂

The sign of cofactor of a₁₁ is (+).

And, cofactor of any element, say a₁₂ is found by eliminating first row and second column.

⇒ Cofactor of a₁₂ = a₂₁ × a₃₃ – a₂₃ × a₃₁

The sign of cofactor of a₁₂ is (-).

Similarly,

First know what the element at position a₁₂ in the matrix is.

In ,

a₁₂ = -3

And as discussed above, the sign at a₁₂ is (-).

For cofactor of -3, eliminate first row and second column in the matrix.

⇒ Cofactor of -3 = (6 × -7) – (4 × 1)

⇒ Cofactor of -3 = -42 – 4

⇒ Cofactor of -3 = -46

Since, the sign of cofactor of -3 is (-), then

Cofactor of -3 = -(-46)

⇒ Cofactor of -3 = 46

Thus, cofactor of -3 is 46.

Question 35.

If find x.

Answer:

9(2x + 5) – 3(5x+2) = 0

⇒ 18x + 45 – 15x – 6 = 0

⇒ 3x + 39 = 0

⇒ 3x = - 39

⇒ x = -13

Question 36.

Find the value of x from the following: .

Answer:

We are given that,

We need to find the value of x.

Determinant of 2 × 2 matrix is found as,

So, determinant of the given matrix is found as,

According to the question, equate this to 0.

2x² – 4 = 0

We need to solve the algebraic equation.

2x²= 4

⇒ x² = 2

Taking square root on both sides of the equation,

⇒ x = ±√2

Hence, the value of x is ±√2.

Question 37.

Write the value of the determinant .

Answer:

We need to find the value of determinant,

Determinant of 3 × 3 matrices is found as,

Similarly,

Thus, the value of .

Question 38.

If |A| = 2, where A is 2 × 2 matrix, find |adj A|.

Answer:

We are given that,

Order of matrix A = 2 × 2

|A| = 2

We need to find the |adj A|.

Let us understand what adjoint of a matrix is.

Let A = [a_ij] be a square matrix of order n × n. Then, the adjoint of the matrix A is transpose of the cofactor of matrix A.

The relationship between adjoint of matrix and determinant of matrix is given as,

|adj A| = |A|^n-1

Where, n = order of the matrix

Putting |A| = 2 in the above equation,

⇒ |adj A| = (2)^n-1 …(i)

Here, order of matrix A = 2

∴, n = 2

Putting n = 2 in equation (i), we get

⇒ |adj A| = (2)^2-1

⇒ |adj A| = (2)¹

⇒ |adj A| = 2

Thus, the |adj A| is 2.

Question 39.

For what is the value of the determinant ?

Answer:

We need to find the value of determinant,

Determinant of 3 × 3 matrices is found as,

Similarly,

Thus, the value of is 8.

Question 40.

For what value of x is the matrix singular?

Answer:

We are given that,

is singular matrix.

We need to find the value of x.

Let us recall the definition of singular matrix.

A singular matrix is a square matrix that doesn’t have a matrix inverse. A matrix ‘A’ is singular iff its determinant is zero, i.e., |A| = 0.

Hence, we just need to find the determinant of the given matrix and equate it to zero.

Determinant of 2 × 2 matrix is found as,

So,

Now, equate this to 0.

That is,

⇒ 3x – 6 = 0

⇒ 3x = 6

⇒ x = 2

Thus, the value of x = 2 for which the matrix is singular.

Question 41.

A matrix A of order 3 × 3 is such that |A| = 4. Find the value of |2A|.

Answer:

We are given that,

Order of matrix A = 3

|A| = 4

We need to find the value of |2A|.

By property of determinant of matrix,

|KA| = Kⁿ|A|

Where, order of the matrix A is n.

Similarly,

|2A| = 2³|A|

[∵, Order of matrix A = 3]

⇒ |2A| = 8|A|

Substituting the value of |A| in the above equation,

⇒ |2A| = 8 × 4

⇒ |2A| = 32

Thus, the value of |2A| is 32.

Question 42.

Evaluate:

Answer:

We need to evaluate the matrix:

Determinant of 2 × 2 matrix is found as,

So,

Using the trigonometric identity,

cos (A + B) = cos A cos B – sin A sin B

Replace A by 15° and B by 75°.

cos (15° + 75°) = cos 15° cos 75° - sin 15° cos 75°

⇒ cos 90° = cos 15° cos 75° - sin 15° cos 75°

So, substituting it, we get

Using the trigonometric identity,

cos 90° = 0

Thus, the value of .

Question 43.

If . Write the cofactor of the element a₃₂.

Answer:

We are given that,

We need to find the cofactor of the element a₃₂.

Let us recall how to find the cofactor of any element:

If we are given with,

Cofactor of any element, say a₁₁ is found by eliminating first row and first column.

⇒ Cofactor of a₁₁ = a₂₂ × a₃₃ – a₂₃ × a₃₂

The sign of cofactor of a₁₁ is (+).

And, cofactor of any element, say a₁₂ is found by eliminating first row and second column.

⇒ Cofactor of a₁₂ = a₂₁ × a₃₃ – a₂₃ × a₃₁

The sign of cofactor of a₁₂ is (-).

So,

In matrix, .

Element at a₃₂ = 2

We need to find the cofactor of 2 at a₃₂.

And as discussed above, the sign at a₃₂ is (-).

For cofactor of a₃₂, eliminate third row and second column in the matrix.

⇒ Cofactor of a₃₂ = 5 × 1 – 8 × 2

⇒ Cofactor of a₃₂ = 5 – 16

⇒ Cofactor of a₃₂ = -11

Since, the sign of cofactor of a₃₂ is (-), then

Cofactor of a₃₂ = -(-11)

⇒ Cofactor of a₃₂ = 11

Thus, cofactor of a₃₂ is 11.

Question 44.

If , then write the value of x.

Answer:

We are given that,

We need to find the value of x.

Determinant of 2 × 2 matrix is found as,

Let us take left hand side (LHS) of the given matrix equation.

⇒ LHS = (x + 1)(x + 2) – (x – 1)(x – 3)

⇒ LHS = (x² + x + 2x + 2) – (x² – x – 3x + 3)

⇒ LHS = (x² + 3x + 2) – (x² – 4x + 3)

⇒ LHS = x² + 3x + 2 – x² + 4x – 3

⇒ LHS = x² – x² + 3x + 4x + 2 – 3

⇒ LHS = 7x – 1

Let us take right hand side (RHS) of the given matrix equation.

⇒ RHS = 4 × 3 – (-1) × 1

⇒ RHS = 12 + 1

⇒ RHS = 13

Now,

LHS = RHS

⇒ 7x – 1 = 13

⇒ 7x = 13 + 1

⇒ 7x = 14

⇒ x = 2

Thus, the value of x is 2.

Question 45.

If , then write the value of x.

Answer:

We are given that,

We need to find the value of x.

Determinant of 2 × 2 matrix is found as,

Let us take left hand side (LHS) of the given matrix equation.

⇒ LHS = 2x(x + 1) – (x + 3)(2(x + 1))

⇒ LHS = (2x² + 2x) – (x + 3)(2x + 2)

⇒ LHS = (2x² + 2x) – (2x² + 2x + 6x + 6)

⇒ LHS = (2x² + 2x) – (2x² + 8x + 6)

⇒ LHS = 2x² + 2x – 2x² – 8x – 6

⇒ LHS = 2x² – 2x² + 2x – 8x – 6

⇒ LHS = -6x – 6

Let us take right hand side (RHS) of the given matrix equation.

⇒ RHS = 1 × 3 – 5 × 3

⇒ RHS = 3 – 15

⇒ RHS = -12

Now,

LHS = RHS

⇒ -6x – 6 = -12

⇒ -6x = -12 + 6

⇒ -6x = -6

⇒ x = 1

Thus, the value of x is 1.

Question 46.

If , find the value of x.

Answer:

We are given that,

We need to find the value of x.

Determinant of 2 × 2 matrix is found as,

Let us take left hand side (LHS) of the given matrix equation.

⇒ LHS = 3x × 4 – 7 × (-2)

⇒ LHS = 12x – (-14)

⇒ LHS = 12x + 14

Let us take right hand side (RHS) of the given matrix equation.

⇒ RHS = 8 × 4 – 7 × 6

⇒ RHS = 32 – 42

⇒ RHS = -10

Now,

LHS = RHS

⇒ 12x + 14 = -10

⇒ 12x = -10 – 14

⇒ 12x = -24

⇒ x = -2

Thus, the value of x is -2.

Question 47.

If , write the value of x.

Answer:

We are given that,

We need to find the value of x.

Determinant of 2 × 2 matrix is found as,

Let us take left hand side (LHS) of the given matrix equation.

⇒ LHS = 2x × x – 5 × 8

⇒ LHS = 2x² – 40

Let us take right hand side (RHS) of the given matrix equation.

⇒ RHS = 6 × 3 – (-2) × 7

⇒ RHS = 18 – (-14)

⇒ RHS = 18 + 14

⇒ RHS = 32

Now,

LHS = RHS

⇒ 2x² – 40 = 32

⇒ 2x² = 32 + 40

⇒ 2x² = 72

⇒ x² = 36

⇒ x = ±√36

⇒ x = ±6

Thus, the value of x is ±6.

Question 48.

If A is a 3 × 3 matrix, |A| ≠ 0 and |3A| = k|A| then write value of k.

Answer:

We are given that,

Order of matrix = 3

|A| ≠ 0

|3A| = k|A|

We need to find the value of k.

In order to find k, we need to solve |3A|.

Using property of determinants,

|kA| = kⁿ|A|

Where, order of A is n × n.

Similarly,

|3A| = 3³|A|

[∵, order of A is 3]

⇒ |3A| = 27|A| …(i)

As, according to the question

|3A| = k|A|

Using (i),

⇒ 27|A| = k|A|

Comparing the left hand side and right hand side, we get

k = 27

Thus, the value of k is 27.

Question 49.

Write the value of the determinant .

Answer:

We need to find the determinant,

Determinant of 2 × 2 matrix is found as,

So,

Using the algebraic identity,

(a + b)(a – b) = a² – b²

Thus, the value of .

Question 50.

Write the value of the determinant .

Answer:

We need to find the value of determinant

Determinant of 3 × 3 matrices is found as,

So,

Re-arranging the equation,

Using the algebraic identity,

(a + b)(a – b) = a² – b²

Thus, the value of is 0.

Question 51.

If , then for any natural number, find the value of Det(Aⁿ).

Answer:

We are given that,

We need to find the det(Aⁿ).

To find det(Aⁿ),

First we need to find Aⁿ, and then take determinant of Aⁿ.

Let us find A².

A² = A.A

Let,

For z₁₁: Dot multiply the first row of the first matrix and first column of the second matrix, then sum up.

That is,

(cos θ, sin θ).(cos θ, -sin θ) = cos θ × cos θ + sin θ × (-sin θ)

⇒ (cos θ, sin θ).(cos θ, -sin θ) = cos² θ – sin² θ

By algebraic identity,

cos 2θ = cos² θ – sin² θ

⇒ (cos θ, sin θ).(cos θ, -sin θ) = cos 2θ

For z₁₂: Dot multiply the first row of the first matrix and second column of the second matrix, then sum up.

That is,

(cos θ, sin θ)(sin θ, cos θ) = cos θ × sin θ + sin θ × cos θ

⇒ (cos θ, sin θ)(sin θ, cos θ) = sin θ cos θ + sin θ cos θ

⇒ (cos θ, sin θ)(sin θ, cos θ) = 2 sin θ cos θ

By algebraic identity,

sin 2θ = 2 sin θ cos θ

⇒ (cos θ, sin θ)(sin θ, cos θ) = sin 2θ

Similarly,

If and , then

Now, taking determinant of Aⁿ,

Determinant of 2 × 2 matrix is found as,

So,

Det(Aⁿ) = cos nθ × cos nθ – sin nθ × (-sin nθ)

⇒ Det(Aⁿ) = cos² nθ + sin² nθ

Using the algebraic identity,

sin² A + cos² A = 1

⇒ Det(Aⁿ) = 1

Thus, Det(Aⁿ) is 1.

Question 52.

Find the maximum value of .

Answer:

We need to find the maximum value of

Let us find the determinant,

Determinant of 3 × 3 matrices is found as,

So,

Multiply and divide by 2 on right hand side,

[∵, By trigonometric identity, sin 2θ = 2 sin θ cos θ]

We need to find the maximum value of .

We know the range of sine function.

-1 ≤ sin A ≤ 1

Or,

-1 ≤ sin 2θ ≤ 1

∴, maximum value of sin 2θ is 1.

⇒ maximum value of = 1/2

Thus, maximum value of

Question 53.

If x ∈ N and , then find the value of x.

Answer:

We are given that,

x ∈ N

We need to find the value of x.

Determinant of 2 × 2 matrix is found as,

So, take

Since,

⇒ 2x² = 8

⇒ x² = 4

⇒ x = ±√4

⇒ x = ±2

Since, x ∈ N

-2 is not a natural number.

Thus, the value of x is 2.

Question 54.

If , write the value of x.

Answer:

We are given that,

We need to find the value of x.

Determinant of 3 × 3 matrices is found as,

So,

Using trigonometric identity,

sin² θ + cos² θ = 1

Since,

⇒ -x³ = 8

⇒ x³ = -8

⇒ x³ = -2 × -2 × -2

Taking cube root on both sides,

⇒ x = -2

Thus, the value of x is -2.

Question 55.

If A is a 3 × 3 matrix, then what will be the value of k if Det(A^-1) = (Det A)^k?

Answer:

We are given that,

Order of matrix = 3 × 3

Det(A^-1) = (Det A)^k

An n-by-n square matrix A is called invertible if there exists an n-by-n square matrix B such that where I_n denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication.

We know that,

If A and B are square matrices of same order, then

Det (AB) = Det (A).Det (B)

Since, A is an invertible matrix, this means that, A has an inverse called A^-1.

Then, if A and A^-1 are inverse matrices, then

Det (AA^-1) = Det (A).Det (A^-1)

By property of inverse matrices,

AA^-1 = I

∴, Det (I) = Det (A).Det (A^-1)

Since, Det (I) = 1

⇒ 1 = Det (A).Det (A^-1)

⇒ Det (A^-1) = Det (A)^-1

Since, according to question,

Det(A^-1) = (Det A)^k

⇒ k = -1

Thus, the value of k is -1.

Determinants

Class 12th Mathematics RD Sharma Volume 1 Solution

Exercise 6.1

Exercise 6.2

Exercise 6.3

Exercise 6.4

Exercise 6.5

Mcq

Very Short Answer

Class 12^th Mathematics RD Sharma Volume 1 Solution