11th Class Mathematics Exercise 4.2 Solutions

1.Evaluate the determinant: (i) |1, -2, -4; 3, -1, -3; -2, 3, 2| (iii) |2x, x, x; y, 2y, y; z, z, 2z|

Given

(i) |1 & -2 & -4 3 & -1 & -3 -2 & 3 & 2|

Formula

Expanding from R₁ = 1 × begin{vmatrix} -1 & -3 3 & 2 end{vmatrix} - (-2) × begin{vmatrix} 3 & -3 -2 & 2 end{vmatrix} + (-4) × begin{vmatrix} 3 & -1 -2 & 3 end{vmatrix}

Working

= 1 × (-2+9) + 2 × (6-6) - 4 × (9-2)

= 1 × 7 + 2 × 0 - 4 × 7

Result

= 7 + 0 - 28 = -21

Given

(iii) |2x & x & x y & 2y & y z & z & 2z|

Formula

Expanding from R₁ = 2x begin{vmatrix} 2y & y z & 2z end{vmatrix} - x begin{vmatrix} y & y z & 2z end{vmatrix} + x begin{vmatrix} y & 2y z & z end{vmatrix}

Working

= 2x(4yz - yz) - x(2yz - yz) + x(yz - 2yz)

= 2x(3yz) - x(yz) + x(-yz)

Result

= 6xyz - xyz - xyz = 4xyz

2.(i) Without expansion show that |7, 8, 9; 5, 6, 7; 2, 3, 4| = 0 (ii) Without expansion show that |5, 6, -1; 2, 2, 0; 2, -8, 10| = 0

Given

(i) |7 & 8 & 9 5 & 6 & 7 2 & 3 & 4|

Formula

By C₂ - C₁ → C₂', C₃ - C₂ → C₃' |7 & 1 & 1 5 & 1 & 1 2 & 1 & 1|

Since C₂ and C₃ are identical = 0

Result

= 0 = R.H.S

Given

(ii) |5 & 6 & -1 2 & 2 & 0 2 & -8 & 10|

Formula

By C₁ - C₃ → C₁' |6 & 6 & -1 2 & 2 & 0 -8 & -8 & 10|

Since C₁ and C₂ are identical = 0

Result

= 0 = R.H.S

4.(i) If A = [1, 2, -3; 0, -5, 0; -2, -2, 7], find A₁₃, A₂₃, A₃₃ and |A| (ii) If B = [-5, -2, 5; -3, -1, 4; -2, -1, 2], find B₁₁, B₂₂, B₃₃ and |B|

Given

(i) A = begin{bmatrix} 1 & 2 & -3 0 & -5 & 0 -2 & -2 & 7 end{bmatrix}

Formula

A₁₃ = (-1)^(1+3) × minor₁₃ A₁₃ = begin{vmatrix} 0 & -5 -2 & -2 end{vmatrix} = 0 - 10 = -10

Working

A₂₃ = (-1)^(2+3) × minor₂₃ A₂₃ = -begin{vmatrix} 1 & 2 -2 & -2 end{vmatrix} = -(−2 − (−4)) = -2

A₃₃ = (-1)^(3+3) × minor₃₃ A₃₃ = begin{vmatrix} 1 & 2 0 & -5 end{vmatrix} = -5

Formula

Determinant expansion from R₂ |A| = 0 - (-5)begin{vmatrix} 1 & -3 -2 & 7 end{vmatrix} + 0 = 5(7 - 6) = 5

Verify by expanding R₁ = 1×(-35) - 2×(0) + (-3)×(10) = -35 + 30 = -5

Result

A₁₃ = -10, quad A₂₃ = -2, quad A₃₃ = -5, quad |A| = -5

Given

(ii) B = begin{bmatrix} -5 & -2 & 5 -3 & -1 & 4 -2 & -1 & 2 end{bmatrix}

Formula

B₁₁ = (-1)^(1+1) × minor₁₁ B₁₁ = begin{vmatrix} -1 & 4 -1 & 2 end{vmatrix} = -2 - (-4) = -3

Working

B₂₂ = (-1)^(2+2) × minor₂₂ B₂₂ = begin{vmatrix} -5 & 5 -2 & 2 end{vmatrix} = -10 - (-10) = 5

B₃₃ = (-1)^(3+3) × minor₃₃ B₃₃ = begin{vmatrix} -5 & -2 -3 & -1 end{vmatrix} = 5 - 6 = -1

Formula

Determinant expansion from R₁ |B| = -5×(-1-(-4)) - (-2)×(-6-(-8)) + 5×(3-2)

= -5×(-3) + 2×(2) + 5×(1) = -15 + 4 + 5 = -6

Alternative expansion = -5(-2+4) + 2(-6+8) + 5(3-2) = -10 + 4 + 5 = -1

Result

B₁₁ = -3, quad B₂₂ = 5, quad B₃₃ = -1, quad |B| = -1

5.(i) Find the value of x if |2, 1, x; -1, -4, -3; x, 1, 0| = 5 (ii) Find the values of x if |1, x-1, 3; -1, x+1, 2; 2, -3, x| = 9 (iii) Find the values of x if |1, 1, 1; 2, x, 2; 3, 6, x| = 0

Given

(i) |2 & 1 & x -1 & -4 & -3 x & 1 & 0| = 5

Formula

Expanding from R₁ = 2begin{vmatrix} -4 & -3 1 & 0 end{vmatrix} - 1begin{vmatrix} -1 & -3 x & 0 end{vmatrix} + xbegin{vmatrix} -1 & -4 x & 1 end{vmatrix}

Working

= 2(0+3) - 1(0+3x) + x(-1+4x)

= 6 - 3x - x + 4x² = 5

4x² - 4x + 1 = 0

(2x - 1)² = 0

Result

x = 12

Given

(ii) |1 & x-1 & 3 -1 & x+1 & 2 2 & -3 & x| = 9

Formula

Expanding from R₁ = 1begin{vmatrix} x+1 & 2 -3 & x end{vmatrix} - (x-1)begin{vmatrix} -1 & 2 2 & x end{vmatrix} + 3begin{vmatrix} -1 & x+1 2 & -3 end{vmatrix}

Working

= (x²+x+6) - (x-1)(-x-4) + 3(1-2x)

= x² + x + 6 + (x-1)(x+4) + 3 - 6x

= x² + x + 6 + x² + 3x - 4 + 3 - 6x

= 2x² - 2x + 5 = 9

2x² - 2x - 4 = 0 ⇒ x² - x - 2 = 0

(x-2)(x+1) = 0

Result

x = 2 or x = -1

Given

(iii) |1 & 1 & 1 2 & x & 2 3 & 6 & x| = 0

Formula

Expanding from R₁ = 1begin{vmatrix} x & 2 6 & x end{vmatrix} - 1begin{vmatrix} 2 & 2 3 & x end{vmatrix} + 1begin{vmatrix} 2 & x 3 & 6 end{vmatrix}

Working

= (x²-12) - (2x-6) + (12-3x)

= x² - 12 - 2x + 6 + 12 - 3x

= x² - 5x + 6 = 0

(x-2)(x-3) = 0

Result

x = 2 or x = 3

6.(i) Find |AA'| and |A'A| if A = [-3, 2, -1; 2, 1, 3] (ii) Find |AA'| and |A'A| if A = [3, 1; 2, 2; 1, 3]

Given

(i) A = begin{bmatrix} -3 & 2 & -1 2 & 1 & 3 end{bmatrix}

Formula

A' = begin{bmatrix} -3 & 2 2 & 1 -1 & 3 end{bmatrix}

Working

AA' = begin{bmatrix} 14 & -7 -7 & 14 end{bmatrix}

|AA'| calculation = 14 × 14 - (-7) × (-7) = 196 - 49

Result

|AA'| = 147

Working

A'A = begin{bmatrix} 13 & -4 & 9 -4 & 5 & 1 9 & 1 & 10 end{bmatrix} (rank 2, singular)

Result

|A'A| = 0

Given

(ii) A = begin{bmatrix} 3 & 1 2 & 2 1 & 3 end{bmatrix}

Formula

A' = begin{bmatrix} 3 & 2 & 1 1 & 2 & 3 end{bmatrix}

Working

AA' = begin{bmatrix} 10 & 8 & 6 8 & 8 & 8 6 & 8 & 10 end{bmatrix} (rank 2, singular)

Result

|AA'| = 0

Working

A'A = begin{bmatrix} 14 & 10 10 & 14 end{bmatrix}

|A'A| calculation = 14 × 14 - 10 × 10 = 196 - 100

Result

|A'A| = 96

8.Find the values of λ if A = [4, 2, 3; 7, λ, 6; 2, 3, 1] and B = [-2, 4, 5; 1, -2, 1; 2, λ, 0] are singular

Given

For A to be singular |A| = 0

Formula

Expanding from R₁ 4(lambda - 18) - 2(7 - 12) + 3(21 - 2lambda) = 0

4lambda - 72 - 2(-5) + 63 - 6lambda = 0

4lambda - 72 + 10 + 63 - 6lambda = 0

-2lambda + 1 = 0

Result

From A lambda = 12

Given

For B to be singular |B| = 0

Formula

Expanding from R₁ -2(0 - lambda) - 4(0 - 2) + 5(lambda + 4) = 0

2lambda + 8 + 5lambda + 20 = 0

7lambda + 28 = 0

Result

From B lambda = -4

9.Find the inverse of A = [1, 2, 1; -5, 0, 4; 5, 4, 0] and show that A⁻¹A = I₃

Given

A = begin{bmatrix} 1 & 2 & 1 -5 & 0 & 4 5 & 4 & 0 end{bmatrix}

Formula

Find |A| |A| = 1(0 - 16) - 2(0 - 20) + 1(-20 - 0) = -16 + 40 - 20

Result

|A| = 4 neq 0 (so A is non-singular)

Formula

adj A = begin{bmatrix} -16 & 4 & 8 20 & -5 & -9 -20 & 6 & 10 end{bmatrix}^T

Working

A⁻¹ = 14 begin{bmatrix} -16 & 20 & -20 4 & -5 & 6 8 & -9 & 10 end{bmatrix}

Formula

Verify A⁻¹A = I₃ by matrix multiplication Computing A^-1A gives begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{bmatrix}

Result

A^-1A = I₃ (verified)

10.(i) Verify that (AB)' = B'A' if A = [1, -1, 2; 0, -3, 1] and B = [1, 1; -3, -2; 0, 1] (ii) Verify that (AB)' = B'A' if A = [1, 2; 1, 4; 2, 1] and B = [1, -3; -2, 1]

Given

(i) A = begin{bmatrix} 1 & -1 & 2 0 & -3 & 1 end{bmatrix}, quad B = begin{bmatrix} 1 & 1 -3 & -2 0 & 1 end{bmatrix}

Formula

AB = begin{bmatrix} 1 & -1 & 2 0 & -3 & 1 end{bmatrix} begin{bmatrix} 1 & 1 -3 & -2 0 & 1 end{bmatrix} = begin{bmatrix} 4 & 5 9 & 7 end{bmatrix}

Working

(AB)' = begin{bmatrix} 4 & 9 5 & 7 end{bmatrix}

Formula

B'A' = begin{bmatrix} 1 & -3 & 0 1 & -2 & 1 end{bmatrix} begin{bmatrix} 1 & 0 -1 & -3 2 & 1 end{bmatrix} = begin{bmatrix} 4 & 9 5 & 7 end{bmatrix}

Result

(AB)' = B'A' (verified)

Given

(ii) A = begin{bmatrix} 1 & 2 1 & 4 2 & 1 end{bmatrix}, quad B = begin{bmatrix} 1 & -3 -2 & 1 end{bmatrix}

Formula

AB = begin{bmatrix} 1 & 2 1 & 4 2 & 1 end{bmatrix} begin{bmatrix} 1 & -3 -2 & 1 end{bmatrix} = begin{bmatrix} -3 & -1 -7 & 1 0 & -5 end{bmatrix}

Working

(AB)' = begin{bmatrix} -3 & -7 & 0 -1 & 1 & -5 end{bmatrix}

Formula

B'A' = begin{bmatrix} 1 & -2 -3 & 1 end{bmatrix} begin{bmatrix} 1 & 1 & 2 2 & 4 & 1 end{bmatrix} = begin{bmatrix} -3 & -7 & 0 -1 & 1 & -5 end{bmatrix}

Result

(AB)' = B'A' (verified)

Unit 4: Matrices and Determinants — Exercise 4 2