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NCERT Solutions · Class 12 Maths · NCERT Mathematics Part I and II Part I · Chapter 4

Chapter 4: Determinants (Algebra)

Step-by-step solutions to all exercises of Chapter 4, Determinants (NCERT Class 12 Maths Part I, 2026-27): Exercise 4.1 (evaluating determinants), 4.2 (area of a triangle, collinearity, equation of a line), 4.3 (minors and cofactors), 4.4 (adjoint and inverse of a matrix), 4.5 (consistency and solving linear equations by the matrix method) and the Miscellaneous Exercise. All 61 questions are answered, with the key answer highlighted.

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Expansion: $∣ A ∣ = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$ , where the cofactor $A_{ij} = (- 1)^{i + j} M_{ij}$ . Area of a triangle: $Δ = \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111$ . $adj A$ is the transpose of the cofactor matrix; $A^{- 1} = \frac{1}{∣ A ∣} adj A$ ( $∣ A ∣ \neq = 0$ ). For $A X = B$ : if $∣ A ∣ \neq = 0$ , $X = A^{- 1} B$ (unique solution); if $∣ A ∣ = 0$ and $(adj A) B \neq = O$ , the system is inconsistent.

Exercise 4.1

Evaluate

2 - 5 4 - 1

Solution

$2 (- 1) - 4 (- 5) = - 2 + 20 = 18$

Evaluate (i)

cos θ sin θ - sin θ cos θ

(ii)

x^{2} - x + 1 x + 1 x - 1 x + 1

Solution

(i) $cos^{2} θ + sin^{2} θ = 1$

(ii) $(x^{2} - x + 1) (x + 1) - (x - 1) (x + 1) = (x^{3} + 1) - (x^{2} - 1) = x^{3} - x^{2} + 2$

(i) 1. (ii) $x^{3} - x^{2} + 2$

A = [1422]

, then show that

∣2 A ∣ = 4∣ A ∣

Solution

$∣ A ∣ = 2 - 8 = - 6$ . $2 A = [2844]$ , so $∣2 A ∣ = 8 - 32 = - 24 = 4 (- 6) = 4∣ A ∣$ .

|2A| = −24 = 4|A|.

A = 100010124

, then show that

∣3 A ∣ = 27∣ A ∣

Solution

A is upper triangular, so $∣ A ∣ = 1 \cdot 1 \cdot 4 = 4$ . $3 A$ is also upper triangular with diagonal 3, 3, 12, so $∣3 A ∣ = 3 \cdot 3 \cdot 12 = 108 = 27 \times 4 = 27∣ A ∣$ .

|3A| = 108 = 27|A|.

Evaluate the determinants (i)

303 - 1 0 - 5 - 2 - 1 0

(ii)

312 - 4 13 5 - 2 1

(iii)

0 - 1 - 2 103 2 - 3 0

(iv)

203 - 1 2 - 5 - 2 - 1 0

Solution

(i) Expand along $R_{2}$ (two zeros): only $a_{23} = - 1$ contributes, with cofactor $(- 1)^{2 + 3} 33 - 1 - 5 = - (- 15 + 3) = 12$ . So $Δ = (- 1) (12) = - 12$ .

(ii) $3 (1 + 6) + 4 (1 + 4) + 5 (3 - 2) = 21 + 20 + 5 = 46$

(iii) $0 - 1 (0 - 6) + 2 (- 3 - 0) = 6 - 6 = 0$

(iv) $2 (0 - 5) + 1 (0 + 3) - 2 (0 - 6) = - 10 + 3 + 12 = 5$

(i) −12. (ii) 46. (iii) 0. (iv) 5.

A = 125114 - 2 - 3 - 9

, find |A|.

Solution

$∣ A ∣ = 1 (- 9 + 12) - 1 (- 18 + 15) - 2 (8 - 5) = 3 + 3 - 6 = 0$

|A| = 0

Find values of x, if (i)

2541 = 2 x 6 4 x

(ii)

2435 = x 2 x 35

Solution

(i) $2 - 20 = 2 x^{2} - 24 \Rightarrow 2 x^{2} = 6 \Rightarrow x = \pm 3$ .

(ii) $10 - 12 = 5 x - 6 x \Rightarrow - 2 = - x \Rightarrow x = 2$ .

(i) $x = \pm 3$ . (ii) x = 2.

x 18 2 x = 61826

, then x is equal to (A) 6 (B) ±6 (C) −6 (D) 0

Solution

$x^{2} - 36 = 36 - 36 = 0 \Rightarrow x = \pm 6$ .

(B)

Exercise 4.2

Find area of the triangle with vertices at the point given in each of the following: (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (−2, −3), (3, 2), (−1, −8)

Solution

(i) $Δ = \frac{1}{2} 164003111 = \frac{1}{2} [1 (0 - 3) - 0 + 1 (18 - 0)] = \frac{15}{2}$ sq units.

(ii) $Δ = \frac{1}{2} ∣2 (1 - 8) - 7 (1 - 10) + 1 (8 - 10) ∣ = \frac{1}{2} ∣ - 14 + 63 - 2∣ = \frac{47}{2}$ sq units.

(iii) Using $Δ = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$ : $Δ = \frac{1}{2} ∣ - 2 (2 + 8) + 3 (- 8 + 3) + (- 1) (- 3 - 2) ∣ = \frac{1}{2} ∣ - 20 - 15 + 5∣ = \frac{1}{2} \cdot 30 = 15$ sq units.

(i) 15/2. (ii) 47/2. (iii) 15 sq units.

Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Solution

$Δ = \frac{1}{2} a b c b + c c + a a + b 111$

Apply $C_{1} \to C_{1} + C_{2}$ : every entry of the first column becomes $a + b + c$ , so $C_{1} = (a + b + c) C_{3}$ . Two proportional columns make the determinant 0. So the area is 0 and the points are collinear.

The determinant is 0 (C₁ + C₂ is a multiple of C₃), so the points are collinear.

Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)

Solution

(i) $\frac{1}{2} ∣ k (0 - 2) - 0 + 1 (8 - 0) ∣ = 4 \Rightarrow ∣ - 2 k + 8∣ = 8 \Rightarrow k = 0$ or $k = 8$ .

(ii) $\frac{1}{2} ∣ - 2 (4 - k) - 0 + 1 (0 - 0) ∣ = 4 \Rightarrow ∣ k - 4∣ = 4 \Rightarrow k = 0$ or $k = 8$ .

(i) k = 0 or 8. (ii) k = 0 or 8.

(i) Find equation of line joining (1, 2) and (3, 6) using determinants. (ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Solution

For a point $(x, y)$ on the line, the three points are collinear, so the area is 0.

(i) $x 13 y 26 111 = x (2 - 6) - y (1 - 3) + (6 - 6) = - 4 x + 2 y = 0 \Rightarrow y = 2 x$ .

(ii) $x 39 y 13 111 = x (1 - 3) - y (3 - 9) + (9 - 9) = - 2 x + 6 y = 0 \Rightarrow x - 3 y = 0$ .

(i) y = 2x. (ii) x − 3y = 0.

If area of triangle is 35 sq units with vertices (2, −6), (5, 4) and (k, 4). Then k is (A) 12 (B) −2 (C) −12, −2 (D) 12, −2

Solution

$\frac{1}{2} 25 k - 6 44 111 = \frac{1}{2} [2 (4 - 4) + 6 (5 - k) + (20 - 4 k)] = \frac{1}{2} (50 - 10 k)$

$∣50 - 10 k ∣ = 70 \Rightarrow k = - 2$ or $k = 12$ .

(D)

Exercise 4.3

Write minors and cofactors of the elements of following determinants: (i)

20 - 4 3

(ii)

a b c d

Solution

(i) $M_{11} = 3$ , $M_{12} = 0$ , $M_{21} = - 4$ , $M_{22} = 2$ ; $A_{11} = 3$ , $A_{12} = 0$ , $A_{21} = 4$ , $A_{22} = 2$ .

(ii) $M_{11} = d$ , $M_{12} = b$ , $M_{21} = c$ , $M_{22} = a$ ; $A_{11} = d$ , $A_{12} = - b$ , $A_{21} = - c$ , $A_{22} = a$ .

As listed above.

Write minors and cofactors of the elements of: (i)

100010001

(ii)

130051 4 - 1 2

Solution

(i) $M_{11} = M_{22} = M_{33} = 1$ and all other minors are 0; the cofactors are the same: $A_{11} = A_{22} = A_{33} = 1$ , the rest 0.

(ii) Minors: $M_{11} = 10 + 1 = 11$ , $M_{12} = 6 - 0 = 6$ , $M_{13} = 3 - 0 = 3$ , $M_{21} = 0 - 4 = - 4$ , $M_{22} = 2 - 0 = 2$ , $M_{23} = 1 - 0 = 1$ , $M_{31} = 0 - 20 = - 20$ , $M_{32} = - 1 - 12 = - 13$ , $M_{33} = 5 - 0 = 5$ .

Cofactors: $A_{11} = 11$ , $A_{12} = - 6$ , $A_{13} = 3$ , $A_{21} = 4$ , $A_{22} = 2$ , $A_{23} = - 1$ , $A_{31} = - 20$ , $A_{32} = 13$ , $A_{33} = 5$ .

As listed above.

Using cofactors of elements of second row, evaluate

Δ = 521302813

Solution

$A_{21} = - (9 - 16) = 7$ , $A_{22} = 15 - 8 = 7$ , $A_{23} = - (10 - 3) = - 7$ .

$Δ = 2 (7) + 0 (7) + 1 (- 7) = 7$

Δ = 7

Using cofactors of elements of third column, evaluate

Δ = 111 x y z y z z x x y

Solution

$A_{13} = z - y$ , $A_{23} = - (z - x)$ , $A_{33} = y - x$ .

$Δ = y z (z - y) - z x (z - x) + x y (y - x) = (x - y) (y - z) (z - x)$

(Expanding: $y z^{2} - y^{2} z - x z^{2} + x^{2} z + x y^{2} - x^{2} y$ , which factorises as above.)

$(x - y) (y - z) (z - x)$

Δ = a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33}

and

A_{ij}

is cofactor of

a_{ij}

, then value of Δ is given by (A)

a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}

(B)

a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}

(C)

a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}

(D)

a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}

Solution

Δ is the sum of the elements of any row or column multiplied by their own cofactors. (D) uses column 1 with its own cofactors.

(D)

Exercise 4.4

Find adjoint of the matrix

[1324]

Solution

For a 2 × 2 matrix, swap the diagonal and change the signs of the other two: $adj A = [4 - 3 - 2 1]$ .

$[4 - 3 - 2 1]$

Find adjoint of the matrix

12 - 2 - 1 30 251

Solution

Cofactors: $A_{11} = 3$ , $A_{12} = - (2 + 10) = - 12$ , $A_{13} = 0 + 6 = 6$ ; $A_{21} = - (- 1 - 0) = 1$ , $A_{22} = 1 + 4 = 5$ , $A_{23} = - (0 - 2) = 2$ ; $A_{31} = - 5 - 6 = - 11$ , $A_{32} = - (5 - 4) = - 1$ , $A_{33} = 3 + 2 = 5$ .

$adj A = 3 - 12 6 152 - 11 - 1 5$

$3 - 12 6 152 - 11 - 1 5$

Verify

A (adj A) = (adj A) A = ∣ A ∣ I

for

[2 - 4 3 - 6]

Solution

$∣ A ∣ = - 12 + 12 = 0$ ; $adj A = [- 6 4 - 3 2]$ .

$A (adj A) = [- 12 + 12 24 - 24 - 6 + 6 12 - 12] = O, (adj A) A = [- 12 + 12 8 - 8 - 18 + 18 12 - 12] = O$

and $∣ A ∣ I = 0 \cdot I = O$ . Verified.

All three equal O since |A| = 0.

Verify

A (adj A) = (adj A) A = ∣ A ∣ I

for

131 - 1 00 2 - 2 3

Solution

$∣ A ∣ = 1 (0) + 1 (9 + 2) + 2 (0) = 11$ .

Cofactors: $A_{11} = 0$ , $A_{12} = - 11$ , $A_{13} = 0$ , $A_{21} = 3$ , $A_{22} = 1$ , $A_{23} = - 1$ , $A_{31} = 2$ , $A_{32} = 8$ , $A_{33} = 3$ , so

$adj A = 0 - 11 0 31 - 1 283$

Multiplying, $A (adj A) = 110001100011 = 11 I$ , and $(adj A) A = 11 I$ as well. Verified.

Both products equal 11I = |A|I.

Find the inverse of

[24 - 2 3]

Solution

$∣ A ∣ = 6 + 8 = 14$ ; $adj A = [3 - 4 22]$ .

$A^{- 1} = \frac{1}{14} [3 - 4 22]$

$\frac{1}{14} [3 - 4 22]$

Find the inverse of

[- 1 - 3 52]

Solution

$∣ A ∣ = - 2 + 15 = 13$ ; $adj A = [23 - 5 - 1]$ .

$A^{- 1} = \frac{1}{13} [23 - 5 - 1]$

$\frac{1}{13} [23 - 5 - 1]$

Find the inverse of

100220345

Solution

$∣ A ∣ = 1 \cdot 2 \cdot 5 = 10$ (triangular).

Cofactors: $A_{11} = 10$ , $A_{12} = 0$ , $A_{13} = 0$ , $A_{21} = - 10$ , $A_{22} = 5$ , $A_{23} = 0$ , $A_{31} = 2$ , $A_{32} = - 4$ , $A_{33} = 2$ .

$A^{- 1} = \frac{1}{10} 1000 - 10 50 2 - 4 2$

$\frac{1}{10} 1000 - 10 50 2 - 4 2$

Find the inverse of

135032 00 - 1

Solution

$∣ A ∣ = 1 \cdot 3 \cdot (- 1) = - 3$ .

Cofactors: $A_{11} = - 3$ , $A_{12} = 3$ , $A_{13} = - 9$ , $A_{21} = 0$ , $A_{22} = - 1$ , $A_{23} = - 2$ , $A_{31} = 0$ , $A_{32} = 0$ , $A_{33} = 3$ .

$A^{- 1} = - \frac{1}{3} - 3 3 - 9 0 - 1 - 2 003$

$- \frac{1}{3} - 3 3 - 9 0 - 1 - 2 003$

Find the inverse of

24 - 7 1 - 1 2 301

Solution

$∣ A ∣ = 2 (- 1 - 0) - 1 (4 - 0) + 3 (8 - 7) = - 2 - 4 + 3 = - 3$ .

Cofactors: $A_{11} = - 1$ , $A_{12} = - 4$ , $A_{13} = 1$ , $A_{21} = 5$ , $A_{22} = 23$ , $A_{23} = - 11$ , $A_{31} = 3$ , $A_{32} = 12$ , $A_{33} = - 6$ .

$A^{- 1} = - \frac{1}{3} - 1 - 4 1 523 - 11 312 - 6$

$- \frac{1}{3} - 1 - 4 1 523 - 11 312 - 6$

Find the inverse of

103 - 1 2 - 2 2 - 3 4

Solution

$∣ A ∣ = 1 (8 - 6) + 1 (0 + 9) + 2 (0 - 6) = 2 + 9 - 12 = - 1$ .

Cofactors: $A_{11} = 2$ , $A_{12} = - 9$ , $A_{13} = - 6$ , $A_{21} = 0$ , $A_{22} = - 2$ , $A_{23} = - 1$ , $A_{31} = - 1$ , $A_{32} = 3$ , $A_{33} = 2$ .

$A^{- 1} = \frac{1}{- 1} 2 - 9 - 6 0 - 2 - 1 - 1 32 = - 2 96 021 1 - 3 - 2$

$- 2 96 021 1 - 3 - 2$

Find the inverse of

100 0 cos α sin α 0 sin α - cos α

Solution

$∣ A ∣ = 1 (- cos^{2} α - sin^{2} α) = - 1$ .

Cofactors: $A_{11} = - 1$ , $A_{12} = A_{13} = A_{21} = A_{31} = 0$ , $A_{22} = - cos α$ , $A_{23} = - sin α$ , $A_{32} = - sin α$ , $A_{33} = cos α$ .

$A^{- 1} = - - 1 00 0 - cos α - sin α 0 - sin α cos α = 100 0 cos α sin α 0 sin α - cos α = A$

$A^{- 1} = A$ (the matrix is its own inverse).

Let

A = [3275]

and

B = [6789]

. Verify that

(A B)^{- 1} = B^{- 1} A^{- 1}

Solution

$A B = [18 + 49 12 + 35 24 + 63 16 + 45] = [67478761]$ , $∣ A B ∣ = 4087 - 4089 = - 2$ :

$(A B)^{- 1} = - \frac{1}{2} [61 - 47 - 87 67]$

$∣ A ∣ = 1$ , $A^{- 1} = [5 - 2 - 7 3]$ ; $∣ B ∣ = - 2$ , $B^{- 1} = - \frac{1}{2} [9 - 7 - 8 6]$ .

$B^{- 1} A^{- 1} = - \frac{1}{2} [45 + 16 - 35 - 12 - 63 - 24 49 + 18] = - \frac{1}{2} [61 - 47 - 87 67] = (A B)^{- 1}$

Both equal $- \frac{1}{2} [61 - 47 - 87 67]$ .

A = [3 - 1 12]

, show that

A^{2} - 5 A + 7 I = O

. Hence find

A^{- 1}

Solution

$A^{2} = [8 - 5 53]$ , so $A^{2} - 5 A + 7 I = [8 - 15 + 7 - 5 + 5 5 - 5 3 - 10 + 7] = O$ .

Multiply by $A^{- 1}$ ( $∣ A ∣ = 7 \neq = 0$ ): $A - 5 I + 7 A^{- 1} = O$ , so

$A^{- 1} = \frac{1}{7} (5 I - A) = \frac{1}{7} [21 - 1 3]$

$A^{- 1} = \frac{1}{7} [21 - 1 3]$

For the matrix

A = [3121]

, find the numbers a and b such that

A^{2} + a A + b I = O

Solution

$A^{2} = [11483]$ .

$A^{2} + a A + b I = [11 + 3 a + b 4 + a 8 + 2 a 3 + a + b] = O$ .

From $4 + a = 0$ : $a = - 4$ . Then $11 - 12 + b = 0 \Rightarrow b = 1$ (check: $3 - 4 + 1 = 0$ ✔).

a = −4, b = 1.

For the matrix

A = 112 12 - 1 1 - 3 3

, show that

A^{3} - 6 A^{2} + 5 A + 11 I = O

. Hence, find

A^{- 1}

Solution

$A^{2} = 4 - 3 7 28 - 3 1 - 14 14, A^{3} = 8 - 23 32 727 - 13 1 - 69 58$

For example, the (1, 1) entry: $8 - 24 + 5 + 11 = 0$ ; all entries work out to 0, so $A^{3} - 6 A^{2} + 5 A + 11 I = O$ .

Multiply by $A^{- 1}$ : $A^{2} - 6 A + 5 I + 11 A^{- 1} = O$ , so

$A^{- 1} = - \frac{1}{11} (A^{2} - 6 A + 5 I) = - \frac{1}{11} 3 - 9 - 5 - 4 13 - 5 41 = \frac{1}{11} - 3 95 4 - 1 - 3 5 - 4 - 1$

$A^{- 1} = \frac{1}{11} - 3 95 4 - 1 - 3 5 - 4 - 1$

A = 2 - 1 1 - 1 2 - 1 1 - 1 2

, verify that

A^{3} - 6 A^{2} + 9 A - 4 I = O

and hence find

A^{- 1}

Solution

$A^{2} = 6 - 5 5 - 5 6 - 5 5 - 5 6, A^{3} = 22 - 21 21 - 21 22 - 21 21 - 21 22$

(1, 1) entry: $22 - 36 + 18 - 4 = 0$ ; (1, 2) entry: $- 21 + 30 - 9 = 0$ ; all entries are 0.

Multiply by $A^{- 1}$ : $A^{2} - 6 A + 9 I - 4 A^{- 1} = O$ , so

$A^{- 1} = \frac{1}{4} (A^{2} - 6 A + 9 I) = \frac{1}{4} 31 - 1 131 - 1 13$

$A^{- 1} = \frac{1}{4} 31 - 1 131 - 1 13$

Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to (A) |A| (B) |A|² (C) |A|³ (D) 3|A|

Solution

$∣ adj A ∣ = ∣ A ∣^{n - 1} = ∣ A ∣^{2}$ for $n = 3$ .

(B)

If A is an invertible matrix of order 2, then det (A⁻¹) is equal to (A) det (A) (B)

\frac{1}{det ( A )}

Solution

$A A^{- 1} = I \Rightarrow ∣ A ∣∣ A^{- 1} ∣ = 1 \Rightarrow ∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ .

(B)

Exercise 4.5

Examine the consistency of the system of equations: x + 2y = 2, 2x + 3y = 3.

Solution

$∣ A ∣ = 1223 = - 1 \neq = 0$ , so a unique solution exists: consistent.

Consistent.

Examine the consistency: 2x − y = 5, x + y = 4.

Solution

$∣ A ∣ = 2 + 1 = 3 \neq = 0$ : consistent.

Consistent.

Examine the consistency: x + 3y = 5, 2x + 6y = 8.

Solution

$∣ A ∣ = 6 - 6 = 0$ . $adj A = [6 - 2 - 3 1]$ and $(adj A) B = [30 - 24 - 10 + 8] = [6 - 2] \neq = O$ : inconsistent.

Inconsistent.

Examine the consistency: x + y + z = 1, 2x + 3y + 2z = 2, ax + ay + 2az = 4.

Solution

$∣ A ∣ = 12 a 13 a 12 2 a = 1 (6 a - 2 a) - 1 (4 a - 2 a) + 1 (2 a - 3 a) = a$

For $a \neq = 0$ , $∣ A ∣ \neq = 0$ , so the system is consistent (unique solution). (If $a = 0$ , the third equation reads $0 = 4$ , which is impossible.)

Consistent (for a ≠ 0).

Examine the consistency: 3x − y − 2z = 2, 2y − z = −1, 3x − 5y = 3.

Solution

$∣ A ∣ = 3 (0 - 5) + 1 (0 + 3) - 2 (0 - 6) = - 15 + 3 + 12 = 0$ .

$adj A = - 5 - 3 - 6 10612536$ , $(adj A) B = - 10 - 10 + 15 - 6 - 6 + 9 - 12 - 12 + 18 = - 5 - 3 - 6 \neq = O$ : inconsistent.

Inconsistent.

Examine the consistency: 5x − y + 4z = 5, 2x + 3y + 5z = 2, 5x − 2y + 6z = −1.

Solution

$∣ A ∣ = 5 (18 + 10) + 1 (12 - 25) + 4 (- 4 - 15) = 140 - 13 - 76 = 51 \neq = 0$ : consistent.

Consistent.

Solve using matrix method: 5x + 2y = 4, 7x + 3y = 5.

Solution

$∣ A ∣ = 15 - 14 = 1$ , $A^{- 1} = [3 - 7 - 2 5]$ .

$X = A^{- 1} B = [3 - 7 - 2 5] [45] = [12 - 10 - 28 + 25] = [2 - 3]$

x = 2, y = −3

Solve using matrix method: 2x − y = −2, 3x + 4y = 3.

Solution

$∣ A ∣ = 8 + 3 = 11$ , $A^{- 1} = \frac{1}{11} [4 - 3 12]$ .

$X = \frac{1}{11} [- 8 + 3 6 + 6] = [- \frac{5}{11} \frac{12}{11}]$

$x = - \frac{5}{11}$ , $y = \frac{12}{11}$

Solve using matrix method: 4x − 3y = 3, 3x − 5y = 7.

Solution

$∣ A ∣ = - 20 + 9 = - 11$ , $A^{- 1} = - \frac{1}{11} [- 5 - 3 34]$ .

$X = - \frac{1}{11} [- 15 + 21 - 9 + 28] = [- \frac{6}{11} - \frac{19}{11}]$

$x = - \frac{6}{11}$ , $y = - \frac{19}{11}$

Solve using matrix method: 5x + 2y = 3, 3x + 2y = 5.

Solution

$∣ A ∣ = 10 - 6 = 4$ , $A^{- 1} = \frac{1}{4} [2 - 3 - 2 5]$ .

$X = \frac{1}{4} [6 - 10 - 9 + 25] = [- 1 4]$

x = −1, y = 4

Solve using matrix method: 2x + y + z = 1, x − 2y − z = 3/2, 3y − 5z = 9.

Solution

$A = 210 1 - 2 3 1 - 1 - 5$ , $∣ A ∣ = 2 (10 + 3) - 1 (- 5 - 0) + 1 (3 - 0) = 26 + 5 + 3 = 34$ .

$adj A = 1353 8 - 10 - 6 13 - 5$ .

$X = \frac{1}{34} 13 + 12 + 9 5 - 15 + 27 3 - 9 - 45 = \frac{1}{34} 3417 - 51 = 1 \frac{1}{2} - \frac{3}{2}$

x = 1, y = 1/2, z = −3/2

Solve using matrix method: x − y + z = 4, 2x + y − 3z = 0, x + y + z = 2.

Solution

$A = 121 - 1 11 1 - 3 1$ , $∣ A ∣ = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1) = 10$ .

$adj A = 4 - 5 1 20 - 2 253$ .

$X = \frac{1}{10} 16 + 0 + 4 - 20 + 0 + 10 4 + 0 + 6 = 2 - 1 1$

x = 2, y = −1, z = 1

Solve using matrix method: 2x + 3y + 3z = 5, x − 2y + z = −4, 3x − y − 2z = 3.

Solution

$A = 213 3 - 2 - 1 31 - 2$ , $∣ A ∣ = 2 (4 + 1) - 3 (- 2 - 3) + 3 (- 1 + 6) = 10 + 15 + 15 = 40$ .

$adj A = 555 3 - 13 11 91 - 7$ .

$X = \frac{1}{40} 25 - 12 + 27 25 + 52 + 3 25 - 44 - 21 = \frac{1}{40} 4080 - 40 = 12 - 1$

x = 1, y = 2, z = −1

Solve using matrix method: x − y + 2z = 7, 3x + 4y − 5z = −5, 2x − y + 3z = 12.

Solution

$A = 132 - 1 4 - 1 2 - 5 3$ , $∣ A ∣ = 1 (12 - 5) + 1 (9 + 10) + 2 (- 3 - 8) = 7 + 19 - 22 = 4$ .

$adj A = 7 - 19 - 11 1 - 1 - 1 - 3 117$ .

$X = \frac{1}{4} 49 - 5 - 36 - 133 + 5 + 132 - 77 + 5 + 84 = \frac{1}{4} 8412 = 213$

x = 2, y = 1, z = 3

A = 231 - 3 21 5 - 4 - 2

, find

A^{- 1}

. Using

A^{- 1}

solve the system of equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y − 2z = −3.

Solution

$∣ A ∣ = 2 (- 4 + 4) + 3 (- 6 + 4) + 5 (3 - 2) = 0 - 6 + 5 = - 1$ .

$adj A = 021 - 1 - 9 - 5 22313$ , so $A^{- 1} = - adj A = 0 - 2 - 1 195 - 2 - 23 - 13$ .

$X = A^{- 1} B = 0 - 5 + 6 - 22 - 45 + 69 - 11 - 25 + 39 = 123$

$A^{- 1} = 0 - 2 - 1 195 - 2 - 23 - 13$ ; x = 1, y = 2, z = 3.

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Solution

Let the costs per kg be x (onion), y (wheat), z (rice): $4 x + 3 y + 2 z = 60$ , $2 x + 4 y + 6 z = 90$ , $6 x + 2 y + 3 z = 70$ .

$∣ A ∣ = 4 (12 - 12) - 3 (6 - 36) + 2 (4 - 24) = 0 + 90 - 40 = 50$ .

$adj A = 030 - 20 - 5 010 10 - 20 10$ .

$X = \frac{1}{50} 0 - 450 + 700 1800 + 0 - 1400 - 1200 + 900 + 700 = \frac{1}{50} 250400400 = 588$

Onion ₹5/kg, wheat ₹8/kg, rice ₹8/kg.

Miscellaneous Exercise on Chapter 4

Prove that the determinant

x - sin θ cos θ sin θ - x 1 cos θ 1 x

is independent of θ.

Solution

$Δ = x (- x^{2} - 1) - sin θ (- x sin θ - cos θ) + cos θ (- sin θ + x cos θ)$

$= - x^{3} - x + x sin^{2} θ + sin θ cos θ - sin θ cos θ + x cos^{2} θ = - x^{3} - x + x = - x^{3}$

This does not contain θ.

Δ = −x³, independent of θ.

Evaluate

cos α cos β - sin β sin α cos β cos α sin β cos β sin α sin β - sin α 0 cos α

Solution

Expanding along $R_{1}$ :

$cos α cos β (cos α cos β) - cos α sin β (- cos α sin β) - sin α (- sin α sin^{2} β - sin α cos^{2} β)$

$= cos^{2} α (cos^{2} β + sin^{2} β) + sin^{2} α (sin^{2} β + cos^{2} β) = cos^{2} α + sin^{2} α = 1$

A^{- 1} = 3 - 15 5 - 1 6 - 2 1 - 5 2

and

B = 1 - 1 0 23 - 2 - 2 01

, find

(A B)^{- 1}

Solution

$∣ B ∣ = 1 (3 - 0) - 2 (- 1 - 0) - 2 (2 - 0) = 3 + 2 - 4 = 1$ , and $B^{- 1} = adj B = 312212625$ .

$(A B)^{- 1} = B^{- 1} A^{- 1} = 312212625 3 - 15 5 - 1 6 - 2 1 - 5 2 = 9 - 2 1 - 3 10 502$

$(A B)^{- 1} = 9 - 2 1 - 3 10 502$

Let

A = 121231115

. Verify that (i)

[adj A]^{- 1} = adj (A^{- 1})

(ii)

(A^{- 1})^{- 1} = A

Solution

$∣ A ∣ = 1 (15 - 1) - 2 (10 - 1) + 1 (2 - 3) = 14 - 18 - 1 = - 5$ .

$adj A = 14 - 9 - 1 - 9 41 - 1 1 - 1$ , $A^{- 1} = - \frac{1}{5} 14 - 9 - 1 - 9 41 - 1 1 - 1$ .

(i) Since $A (adj A) = ∣ A ∣ I$ , $(adj A)^{- 1} = \frac{A}{∣ A ∣} = - \frac{1}{5} 121231115$ . Also $adj (A^{- 1}) = ∣ A^{- 1} ∣ (A^{- 1})^{- 1} = \frac{1}{∣ A ∣} A$ , which is the same matrix. So the two sides are equal.

(ii) $∣ A^{- 1} ∣ = - \frac{1}{5}$ and computing the inverse of $A^{- 1}$ by the adjoint method gives back $121231115 = A$ (as $A^{- 1} A = A A^{- 1} = I$ ).

Both sides of (i) equal $- \frac{1}{5} A$ , and $(A^{- 1})^{- 1} = A$ .

Evaluate

x y x + y y x + y x x + y x y

Solution

$R_{1} \to R_{1} + R_{2} + R_{3}$ : every entry of $R_{1}$ becomes $2 (x + y)$ :

$Δ = 2 (x + y) 1 y x + y 1 x + y x 1 x y$

$C_{2} \to C_{2} - C_{1}$ , $C_{3} \to C_{3} - C_{1}$ :

$= 2 (x + y) 1 y x + y 0 x - y 0 x - y - x = 2 (x + y) [- x^{2} + y (x - y)] = 2 (x + y) (- x^{2} + x y - y^{2})$

$= - 2 (x + y) (x^{2} - x y + y^{2}) = - 2 (x^{3} + y^{3})$

$- 2 (x^{3} + y^{3})$

Evaluate

111 x x + y x y y x + y

Solution

$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ :

$100 x y 0 y 0 x = 1 (x y - 0) = x y$

Solve the system of equations

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4

\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1

\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2

Solution

Put $u = \frac{1}{x}$ , $v = \frac{1}{y}$ , $w = \frac{1}{z}$ : $2 u + 3 v + 10 w = 4$ , $4 u - 6 v + 5 w = 1$ , $6 u + 9 v - 20 w = 2$ .

$∣ A ∣ = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) = 150 + 330 + 720 = 1200$ .

$adj A = 7511072 150 - 100 0 7530 - 24$ .

$u v w = \frac{1}{1200} 300 + 150 + 150 440 - 100 + 60 288 + 0 - 48 = \frac{1}{1200} 600400240 = \frac{1}{2} \frac{1}{3} \frac{1}{5}$

So $x = 2$ , $y = 3$ , $z = 5$ .

x = 2, y = 3, z = 5

If x, y, z are nonzero real numbers, then the inverse of matrix

A = x 00 0 y 0 00 z

is (A)

x^{- 1} 00 0 y^{- 1} 0 00 z^{- 1}

(B)

x y z x^{- 1} 00 0 y^{- 1} 0 00 z^{- 1}

(C)

\frac{1}{x y z} x 00 0 y 0 00 z

(D)

\frac{1}{x y z} 100010001

Solution

$∣ A ∣ = x y z$ and $adj A = diag (y z, x z, x y)$ , so $A^{- 1} = diag (\frac{1}{x}, \frac{1}{y}, \frac{1}{z})$ .

(A)

Let

A = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1

, where

0 \leq θ \leq 2 π

. Then (A) Det(A) = 0 (B) Det(A) ∈ (2, ∞) (C) Det(A) ∈ (2, 4) (D) Det(A) ∈ [2, 4]

Solution

$det A = 1 (1 + sin^{2} θ) - sin θ (- sin θ + sin θ) + 1 (sin^{2} θ + 1) = 2 + 2 sin^{2} θ$

Since $0 \leq sin^{2} θ \leq 1$ , $det A \in [2, 4]$ .

(D)

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