Chapter 7 of 15

Updated 6 August 2026

NCERT Solutions

Determinants — NCERT Solutions

Madhya Pradesh Board · Class 12 · Mathematics

NCERT Solutions for Determinants, Madhya Pradesh Board Class 12 Mathematics: 72 textbook questions solved step by step.

111 questions48 flashcards15 formulas & key relations5 concepts

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72 Questions Solved · 6 Sections

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Exercise 4.1

1Evaluate the determinant

\left|\begin{array}{cc}2 & 4\\-5 & -1\end{array}\right|

Show solution

Given: $\Delta = \left|\begin{array}{cc}2 & 4\\-5 & -1\end{array}\right|$

Formula: For a $2\times2$ determinant, $\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = ad - bc$

Working:
$\Delta = (2)(-1) - (4)(-5) = -2 + 20 = 18$

Answer: $\Delta = 18$

2(i)Evaluate

\left|\begin{array}{cc}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{array}\right|

Show solution

Given: $\Delta = \left|\begin{array}{cc}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{array}\right|$

Working:
$\Delta = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)$
$= \cos^2\theta + \sin^2\theta = 1$

Answer: $\Delta = 1$

2(ii)Evaluate

\left|\begin{array}{cc}x^2-x+1 & x-1\\x+1 & x+1\end{array}\right|

Show solution

Given: $\Delta = \left|\begin{array}{cc}x^2-x+1 & x-1\\x+1 & x+1\end{array}\right|$

Working:
$\Delta = (x^2-x+1)(x+1) - (x-1)(x+1)$
$= (x^3 + x^2 - x^2 - x + x + 1) - (x^2 - 1)$
$= (x^3 + 1) - (x^2 - 1)$
$= x^3 + 1 - x^2 + 1$
$= x^3 - x^2 + 2$

Answer: $\Delta = x^3 - x^2 + 2$

3If

A = \begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}

, then show that

|2A| = 4|A|

.Show solution

Given: $A = \begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}$

Step 1: Find $|A|$ .
$|A| = (1)(2) - (2)(4) = 2 - 8 = -6$

Step 2: Find $2A$ .
$2A = \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}$

Step 3: Find $|2A|$ .
$|2A| = (2)(4) - (4)(8) = 8 - 32 = -24$

Step 4: Verify.
$4|A| = 4 \times (-6) = -24 = |2A| \quad \checkmark$

Hence $|2A| = 4|A|$ is proved.

4If

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

, then show that

|3A| = 27|A|

.Show solution

Given: $A = \begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}$

Step 1: Find $|A|$ (expanding along $R_1$ ).
$|A| = 1\left|\begin{array}{cc}1&2\\0&4\end{array}\right| - 0 + 1\left|\begin{array}{cc}0&1\\0&0\end{array}\right|$
$= 1(4-0) - 0 + 1(0-0) = 4$

Step 2: Find $3A$ .
$3A = \begin{bmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 & 12\end{bmatrix}$

Step 3: Find $|3A|$ (expanding along $R_1$ ).
$|3A| = 3\left|\begin{array}{cc}3&6\\0&12\end{array}\right| - 0 + 3\left|\begin{array}{cc}0&3\\0&0\end{array}\right|$
$= 3(36-0) - 0 + 3(0-0) = 108$

Step 4: Verify.
$27|A| = 27 \times 4 = 108 = |3A| \quad \checkmark$

Hence $|3A| = 27|A|$ is proved. (Note: For an $n\times n$ matrix, $|kA| = k^n|A|$ ; here $n=3$ , so $|3A|=3^3|A|=27|A|$ .)

5(i)Evaluate

\left|\begin{array}{ccc}3 & -1 & -2\\0 & 0 & -1\\3 & -5 & 0\end{array}\right|

Show solution

Expanding along $R_2$ (since it has two zeros):
$\Delta = -0\cdot M_{21} + 0\cdot M_{22} - (-1)\cdot M_{23}$
$= 0 - 0 + 1\cdot\left|\begin{array}{cc}3&-1\\3&-5\end{array}\right|$
$= 1\cdot[(3)(-5)-(-1)(3)]$
$= (-15+3) = -12$

Answer: $\Delta = -12$

5(ii)Evaluate

\left|\begin{array}{ccc}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{array}\right|

Show solution

Expanding along $R_1$ :
$\Delta = 3\left|\begin{array}{cc}1&-2\\3&1\end{array}\right| -(-4)\left|\begin{array}{cc}1&-2\\2&1\end{array}\right| +5\left|\begin{array}{cc}1&1\\2&3\end{array}\right|$
$= 3(1+6) + 4(1+4) + 5(3-2)$
$= 3(7) + 4(5) + 5(1)$
$= 21 + 20 + 5 = 46$

Answer: $\Delta = 46$

5(iii)Evaluate

\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}

Show solution

Expanding along $R_1$ :
$\Delta = 0\left|\begin{array}{cc}0&-3\\3&0\end{array}\right| -1\left|\begin{array}{cc}-1&-3\\-2&0\end{array}\right| +2\left|\begin{array}{cc}-1&0\\-2&3\end{array}\right|$
$= 0 - 1[(−1)(0)−(−3)(−2)] + 2[(−1)(3)−(0)(−2)]$
$= 0 - 1[0 - 6] + 2[-3 - 0]$
$= 0 + 6 - 6 = 0$

Answer: $\Delta = 0$

5(iv)Evaluate

\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}

Show solution

Expanding along $R_1$ :
$\Delta = 2\left|\begin{array}{cc}2&-1\\-5&0\end{array}\right| -(-1)\left|\begin{array}{cc}0&-1\\3&0\end{array}\right| +(-2)\left|\begin{array}{cc}0&2\\3&-5\end{array}\right|$
$= 2[(2)(0)-(-1)(-5)] + 1[(0)(0)-(-1)(3)] + (-2)[(0)(-5)-(2)(3)]$
$= 2[0-5] + 1[0+3] + (-2)[0-6]$
$= 2(-5) + 3 + (-2)(-6)$
$= -10 + 3 + 12 = 5$

Answer: $\Delta = 5$

6If

A = \begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}

, find

|A|

.Show solution

Expanding along $R_1$ :
$|A| = 1\left|\begin{array}{cc}1&-3\\4&-9\end{array}\right| -1\left|\begin{array}{cc}2&-3\\5&-9\end{array}\right| +(-2)\left|\begin{array}{cc}2&1\\5&4\end{array}\right|$
$= 1[(-9)-(-12)] - 1[(-18)-(-15)] + (-2)[(8)-(5)]$
$= 1[3] - 1[-3] + (-2)[3]$
$= 3 + 3 - 6 = 0$

Answer: $|A| = 0$

7(i)Find values of

x

\begin{vmatrix}2 & 4\\5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4\\6 & x\end{vmatrix}

Show solution

LHS:
$\begin{vmatrix}2&4\\5&1\end{vmatrix} = (2)(1)-(4)(5) = 2-20 = -18$

RHS:
$\begin{vmatrix}2x&4\\6&x\end{vmatrix} = (2x)(x)-(4)(6) = 2x^2 - 24$

Setting LHS = RHS:
$2x^2 - 24 = -18$
$2x^2 = 6$
$x^2 = 3$
$x = \pm\sqrt{3}$

Answer: $x = \pm\sqrt{3}$

7(ii)Find values of

x

\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix} = \begin{vmatrix}x & 3\\2x & 5\end{vmatrix}

Show solution

LHS:
$\begin{vmatrix}2&3\\4&5\end{vmatrix} = (2)(5)-(3)(4) = 10-12 = -2$

RHS:
$\begin{vmatrix}x&3\\2x&5\end{vmatrix} = (x)(5)-(3)(2x) = 5x - 6x = -x$

Setting LHS = RHS:
$-x = -2$
$x = 2$

Answer: $x = 2$

8If

\begin{vmatrix}x & 2\\18 & x\end{vmatrix} = \begin{vmatrix}6 & 2\\18 & 6\end{vmatrix}

, then

x

is equal to: (A) 6, (B)

\pm 6

, (C)

-6

, (D) 0Show solution

Correct Option: (B) $\pm 6$

LHS: $x^2 - 36$

RHS: $36 - 36 = 0$

$x^2 - 36 = 0 \implies x^2 = 36 \implies x = \pm 6$

Hence the correct answer is (B) $\pm 6$ .

Exercise 4.2

1(i)Find area of the triangle with vertices

(1,0)

(6,0)

(4,3)

.Show solution

Formula:
$\Delta = \frac{1}{2}\left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right|$

Working:
$\Delta = \frac{1}{2}\left|\begin{array}{ccc}1&0&1\\6&0&1\\4&3&1\end{array}\right|$

Expanding along $C_2$ (or $R_1$ ):
$= \frac{1}{2}\left[1(0\cdot1 - 1\cdot3) - 0 + 1(6\cdot3 - 0\cdot4)\right]$
$= \frac{1}{2}\left[1(0-3) - 0 + 1(18-0)\right]$
$= \frac{1}{2}[-3 + 18] = \frac{15}{2}$

Answer: Area $= \dfrac{15}{2}$ sq. units

1(ii)Find area of the triangle with vertices

(2,7)

(1,1)

(10,8)

.Show solution

$\Delta = \frac{1}{2}\left|\begin{array}{ccc}2&7&1\\1&1&1\\10&8&1\end{array}\right|$

Expanding along $R_1$ :
$= \frac{1}{2}\left[2(1-8) - 7(1-10) + 1(8-10)\right]$
$= \frac{1}{2}\left[2(-7) - 7(-9) + 1(-2)\right]$
$= \frac{1}{2}[-14 + 63 - 2]$
$= \frac{47}{2}$

Answer: Area $= \dfrac{47}{2}$ sq. units

1(iii)Find area of the triangle with vertices

(-2,-3)

(3,2)

(-1,-8)

.Show solution

$\Delta = \frac{1}{2}\left|\begin{array}{ccc}-2&-3&1\\3&2&1\\-1&-8&1\end{array}\right|$

Expanding along $R_1$ :
$= \frac{1}{2}\left[(-2)(2\cdot1 - 1\cdot(-8)) - (-3)(3\cdot1 - 1\cdot(-1)) + 1(3\cdot(-8) - 2\cdot(-1))\right]$
$= \frac{1}{2}\left[(-2)(2+8) + 3(3+1) + 1(-24+2)\right]$
$= \frac{1}{2}\left[-20 + 12 - 22\right]$
$= \frac{1}{2}(-30) = -15$

Since area is positive:

Answer: Area $= |-15| = 15$ sq. units

2Show that points A

(a, b+c)

, B

(b, c+a)

, C

(c, a+b)

are collinear.Show solution

To show: Area of triangle ABC = 0.

$\text{Area} = \frac{1}{2}\left|\begin{array}{ccc}a&b+c&1\\b&c+a&1\\c&a+b&1\end{array}\right|$

Observe: In each row, sum of first two elements equals $a+b+c$ :

Row 1: $a + (b+c) = a+b+c$
Row 2: $b + (c+a) = a+b+c$
Row 3: $c + (a+b) = a+b+c$

Apply $C_2 \to C_1 + C_2$ :
$= \frac{1}{2}\left|\begin{array}{ccc}a&a+b+c&1\\b&a+b+c&1\\c&a+b+c&1\end{array}\right|$

Take $(a+b+c)$ common from $C_2$ :
$= \frac{a+b+c}{2}\left|\begin{array}{ccc}a&1&1\\b&1&1\\c&1&1\end{array}\right|$

Since $C_2 = C_3$ , the determinant $= 0$ .

Hence Area $= 0$ , so A, B, C are collinear. $\blacksquare$

3(i)Find values of

k

if area of triangle is 4 sq. units and vertices are

(k,0)

(4,0)

(0,2)

.Show solution

Given: Area = 4 sq. units.

$\frac{1}{2}\left|\begin{array}{ccc}k&0&1\\4&0&1\\0&2&1\end{array}\right| = \pm 4$

Expanding along $C_2$ :
$\frac{1}{2}\left[-0 + 0 - 2\left|\begin{array}{cc}k&1\\4&1\end{array}\right|\right] = \pm 4$

$\frac{1}{2}\cdot(-2)(k-4) = \pm 4$

$-(k-4) = \pm 4$

Case 1: $-(k-4) = 4 \Rightarrow k-4 = -4 \Rightarrow k = 0$

Case 2: $-(k-4) = -4 \Rightarrow k-4 = 4 \Rightarrow k = 8$

Answer: $k = 0$ or $k = 8$

3(ii)Find values of

k

if area of triangle is 4 sq. units and vertices are

(-2,0)

(0,4)

(0,k)

.Show solution

Given: Area = 4 sq. units.

$\frac{1}{2}\left|\begin{array}{ccc}-2&0&1\\0&4&1\\0&k&1\end{array}\right| = \pm 4$

Expanding along $C_1$ :
$\frac{1}{2}\left[(-2)\left|\begin{array}{cc}4&1\\k&1\end{array}\right|\right] = \pm 4$

$\frac{1}{2}\cdot(-2)(4-k) = \pm 4$

$-(4-k) = \pm 4$

Case 1: $-(4-k) = 4 \Rightarrow k-4 = 4 \Rightarrow k = 8$

Case 2: $-(4-k) = -4 \Rightarrow k-4 = -4 \Rightarrow k = 0$

Answer: $k = 0$ or $k = 8$

4(i)Find equation of line joining

(1,2)

and

(3,6)

using determinants.Show solution

Let $P(x,y)$ be any point on the line. Then A $(1,2)$ , B $(3,6)$ , P $(x,y)$ are collinear, so area of triangle ABP = 0.

$\frac{1}{2}\left|\begin{array}{ccc}1&2&1\\3&6&1\\x&y&1\end{array}\right| = 0$

Expanding along $R_1$ :
$1(6-y) - 2(3-x) + 1(3y-6x) = 0$
$6 - y - 6 + 2x + 3y - 6x = 0$
$2y - 4x = 0$
$y = 2x$

Answer: Equation of line is $y = 2x$ .

4(ii)Find equation of line joining

(3,1)

and

(9,3)

using determinants.Show solution

Let $P(x,y)$ be any point on the line. Then the three points are collinear:

$\frac{1}{2}\left|\begin{array}{ccc}3&1&1\\9&3&1\\x&y&1\end{array}\right| = 0$

Expanding along $R_1$ :
$3(3-y) - 1(9-x) + 1(9y-3x) = 0$
$9 - 3y - 9 + x + 9y - 3x = 0$
$6y - 2x = 0$
$x - 3y = 0$

Answer: Equation of line is $x = 3y$ (or $x - 3y = 0$ ).

5If 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

Show solution

Correct Option: (D) $12, -2$

$\frac{1}{2}\left|\begin{array}{ccc}2&-6&1\\5&4&1\\k&4&1\end{array}\right| = \pm 35$

Expanding along $R_1$ :
$\frac{1}{2}\left[2(4-4)-(-6)(5-k)+1(20-4k)\right] = \pm 35$
$\frac{1}{2}\left[0 + 6(5-k) + 20 - 4k\right] = \pm 35$
$\frac{1}{2}\left[30 - 6k + 20 - 4k\right] = \pm 35$
$\frac{1}{2}(50 - 10k) = \pm 35$
$50 - 10k = \pm 70$

Case 1: $50 - 10k = 70 \Rightarrow -10k = 20 \Rightarrow k = -2$

Case 2: $50 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12$

Answer: $k = 12$ or $k = -2$ , i.e., option (D).

Exercise 4.3

1(i)Write Minors and Cofactors of the elements of

\left|\begin{array}{cc}2&-4\\0&3\end{array}\right|

Show solution

The elements are $a_{11}=2,\ a_{12}=-4,\ a_{21}=0,\ a_{22}=3$ .

Minors:

$M_{11}$ = determinant after deleting $R_1, C_1$ $= 3$
$M_{12}$ = determinant after deleting $R_1, C_2$ $= 0$
$M_{21}$ = determinant after deleting $R_2, C_1$ $= -4$
$M_{22}$ = determinant after deleting $R_2, C_2$ $= 2$

Cofactors ( $A_{ij} = (-1)^{i+j}M_{ij}$ ):

$A_{11} = (-1)^{1+1}M_{11} = +3 = 3$
$A_{12} = (-1)^{1+2}M_{12} = -0 = 0$
$A_{21} = (-1)^{2+1}M_{21} = -(-4) = 4$
$A_{22} = (-1)^{2+2}M_{22} = +2 = 2$

1(ii)Write Minors and Cofactors of the elements of

\left|\begin{array}{cc}a&c\\b&d\end{array}\right|

Show solution

The elements are $a_{11}=a,\ a_{12}=c,\ a_{21}=b,\ a_{22}=d$ .

Minors:

$M_{11} = d$
$M_{12} = b$
$M_{21} = c$
$M_{22} = a$

Cofactors:

$A_{11} = (-1)^{2}d = d$
$A_{12} = (-1)^{3}b = -b$
$A_{21} = (-1)^{3}c = -c$
$A_{22} = (-1)^{4}a = a$

2(i)Write Minors and Cofactors of the elements of

\left|\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right|

Show solution

This is the identity matrix $I_3$ .

Minors (each $M_{ij}$ is the determinant of the $2\times2$ submatrix after deleting row $i$ and column $j$ ):

$M_{11}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1,\quad M_{12}=\left|\begin{array}{cc}0&0\\0&1\end{array}\right|=0,\quad M_{13}=\left|\begin{array}{cc}0&1\\0&0\end{array}\right|=0$
$M_{21}=\left|\begin{array}{cc}0&0\\0&1\end{array}\right|=0,\quad M_{22}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1,\quad M_{23}=\left|\begin{array}{cc}1&0\\0&0\end{array}\right|=0$
$M_{31}=\left|\begin{array}{cc}0&0\\1&0\end{array}\right|=0,\quad M_{32}=\left|\begin{array}{cc}1&0\\0&0\end{array}\right|=0,\quad M_{33}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1$

Cofactors ( $A_{ij}=(-1)^{i+j}M_{ij}$ ):
$A_{11}=1,\ A_{12}=0,\ A_{13}=0$
$A_{21}=0,\ A_{22}=1,\ A_{23}=0$
$A_{31}=0,\ A_{32}=0,\ A_{33}=1$

2(ii)Write Minors and Cofactors of the elements of

\left|\begin{array}{ccc}1&0&4\\3&5&-1\\0&1&2\end{array}\right|

Show solution

Minors:
$M_{11}=\left|\begin{array}{cc}5&-1\\1&2\end{array}\right|=10+1=11$
$M_{12}=\left|\begin{array}{cc}3&-1\\0&2\end{array}\right|=6-0=6$
$M_{13}=\left|\begin{array}{cc}3&5\\0&1\end{array}\right|=3-0=3$
$M_{21}=\left|\begin{array}{cc}0&4\\1&2\end{array}\right|=0-4=-4$
$M_{22}=\left|\begin{array}{cc}1&4\\0&2\end{array}\right|=2-0=2$
$M_{23}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1-0=1$
$M_{31}=\left|\begin{array}{cc}0&4\\5&-1\end{array}\right|=0-20=-20$
$M_{32}=\left|\begin{array}{cc}1&4\\3&-1\end{array}\right|=-1-12=-13$
$M_{33}=\left|\begin{array}{cc}1&0\\3&5\end{array}\right|=5-0=5$

Cofactors ( $A_{ij}=(-1)^{i+j}M_{ij}$ ):
$A_{11}=+11=11,\quad A_{12}=-6,\quad A_{13}=+3=3$
$A_{21}=-(-4)=4,\quad A_{22}=+2=2,\quad A_{23}=-1$
$A_{31}=+(-20)=-20,\quad A_{32}=-(-13)=13,\quad A_{33}=+5=5$

3Using Cofactors of elements of second row, evaluate

\Delta = \left|\begin{array}{ccc}5&3&8\\2&0&1\\1&2&3\end{array}\right|

Show solution

Elements of second row: $a_{21}=2,\ a_{22}=0,\ a_{23}=1$

Cofactors of second row elements:
$A_{21} = (-1)^{2+1}\left|\begin{array}{cc}3&8\\2&3\end{array}\right| = -(9-16) = -(-7) = 7$
$A_{22} = (-1)^{2+2}\left|\begin{array}{cc}5&8\\1&3\end{array}\right| = +(15-8) = 7$
$A_{23} = (-1)^{2+3}\left|\begin{array}{cc}5&3\\1&2\end{array}\right| = -(10-3) = -7$

Expanding along $R_2$ :
$\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$
$= 2(7) + 0(7) + 1(-7)$
$= 14 + 0 - 7 = 7$

Answer: $\Delta = 7$

4Using Cofactors of elements of third column, evaluate

\Delta = \left|\begin{array}{ccc}1&x&yz\\1&y&zx\\1&z&xy\end{array}\right|

Show solution

Elements of third column: $a_{13}=yz,\ a_{23}=zx,\ a_{33}=xy$

Cofactors:
$A_{13} = (-1)^{1+3}\left|\begin{array}{cc}1&y\\1&z\end{array}\right| = +(z-y)$
$A_{23} = (-1)^{2+3}\left|\begin{array}{cc}1&x\\1&z\end{array}\right| = -(z-x) = (x-z)$
$A_{33} = (-1)^{3+3}\left|\begin{array}{cc}1&x\\1&y\end{array}\right| = +(y-x)$

Expanding along $C_3$ :
$\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$
$= yz(z-y) + zx(x-z) + xy(y-x)$
$= yz^2 - y^2z + x^2z - xz^2 + xy^2 - x^2y$
$= x^2(z-y) + y^2(x-z) + z^2(y-x)$
$= (x-y)(y-z)(z-x)$

Answer: $\Delta = (x-y)(y-z)(z-x)$

5If

\Delta = \left|\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right|

and

A_{ij}

is Cofactor of

a_{ij}

, then value of

\Delta

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}

Show solution

Correct Option: (D) $a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}$

The value of a determinant is obtained by multiplying elements of a row (or column) with their corresponding cofactors and summing. Option (D) represents expansion along the first column: each element $a_{11}, a_{21}, a_{31}$ of column 1 is multiplied by its own cofactor $A_{11}, A_{21}, A_{31}$ respectively. This is a valid expansion.

(A) mixes row 1 elements with row 3 cofactors — gives 0, not $\Delta$ .
(B) mixes row 1 elements with column 1 cofactors — incorrect pairing.
(C) mixes row 2 elements with row 1 cofactors — gives 0, not $\Delta$ .

Answer: (D)

Exercise 4.4

1Find adjoint of the matrix

\begin{bmatrix}1&2\\3&4\end{bmatrix}

Show solution

Given: $A = \begin{bmatrix}1&2\\3&4\end{bmatrix}$

Cofactors:
$A_{11}=+4,\quad A_{12}=-3,\quad A_{21}=-2,\quad A_{22}=+1$

Adjoint = Transpose of cofactor matrix:
$\text{adj}(A) = \begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix} = \begin{bmatrix}4&-2\\-3&1\end{bmatrix}$

Answer: $\text{adj}(A) = \begin{bmatrix}4&-2\\-3&1\end{bmatrix}$

2Find adjoint of the matrix

\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}

Show solution

Given: $A = \begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}$

Cofactors:
$A_{11}=+\left|\begin{array}{cc}3&5\\0&1\end{array}\right|=3,\quad A_{12}=-\left|\begin{array}{cc}2&5\\-2&1\end{array}\right|=-(2+10)=-12$
$A_{13}=+\left|\begin{array}{cc}2&3\\-2&0\end{array}\right|=(0+6)=6$
$A_{21}=-\left|\begin{array}{cc}-1&2\\0&1\end{array}\right|=-(-1-0)=1,\quad A_{22}=+\left|\begin{array}{cc}1&2\\-2&1\end{array}\right|=(1+4)=5$
$A_{23}=-\left|\begin{array}{cc}1&-1\\-2&0\end{array}\right|=-(0-2)=2$
$A_{31}=+\left|\begin{array}{cc}-1&2\\3&5\end{array}\right|=(-5-6)=-11,\quad A_{32}=-\left|\begin{array}{cc}1&2\\2&5\end{array}\right|=-(5-4)=-1$
$A_{33}=+\left|\begin{array}{cc}1&-1\\2&3\end{array}\right|=(3+2)=5$

Adjoint = Transpose of cofactor matrix:
$\text{adj}(A) = \begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix} = \begin{bmatrix}3&1&-11\\-12&5&-1\\6&2&5\end{bmatrix}$

3Verify

A(\text{adj}A) = (\text{adj}A)A = |A|I

for

A = \begin{bmatrix}2&3\\-4&-6\end{bmatrix}

Show solution

Step 1: $|A| = (2)(-6)-(3)(-4) = -12+12 = 0$

So $|A|I = 0\cdot I = \begin{bmatrix}0&0\\0&0\end{bmatrix}$

Step 2: Cofactors: $A_{11}=-6,\ A_{12}=4,\ A_{21}=-3,\ A_{22}=2$

$\text{adj}(A) = \begin{bmatrix}-6&-3\\4&2\end{bmatrix}$

Step 3: Compute $A(\text{adj}A)$ :
$\begin{bmatrix}2&3\\-4&-6\end{bmatrix}\begin{bmatrix}-6&-3\\4&2\end{bmatrix} = \begin{bmatrix}-12+12&-6+6\\24-24&12-12\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$

Step 4: Compute $(\text{adj}A)A$ :
$\begin{bmatrix}-6&-3\\4&2\end{bmatrix}\begin{bmatrix}2&3\\-4&-6\end{bmatrix} = \begin{bmatrix}-12+12&-18+18\\8-8&12-12\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$

Hence $A(\text{adj}A) = (\text{adj}A)A = |A|I = \begin{bmatrix}0&0\\0&0\end{bmatrix}$ . $\checkmark$

4Verify

A(\text{adj}A) = (\text{adj}A)A = |A|I

for

A = \begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}

Show solution

Step 1: Find $|A|$ (expanding along $R_1$ ):
$|A| = 1(0-0)-(-1)(9+2)+2(0-0) = 0+11+0 = 11$

So $|A|I = 11I = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}$

Step 2: Find cofactors:
$A_{11}=\left|\begin{array}{cc}0&-2\\0&3\end{array}\right|=0,\quad A_{12}=-\left|\begin{array}{cc}3&-2\\1&3\end{array}\right|=-(9+2)=-11$
$A_{13}=\left|\begin{array}{cc}3&0\\1&0\end{array}\right|=0$
$A_{21}=-\left|\begin{array}{cc}-1&2\\0&3\end{array}\right|=-(-3-0)=3,\quad A_{22}=\left|\begin{array}{cc}1&2\\1&3\end{array}\right|=3-2=1$
$A_{23}=-\left|\begin{array}{cc}1&-1\\1&0\end{array}\right|=-(0+1)=-1$
$A_{31}=\left|\begin{array}{cc}-1&2\\0&-2\end{array}\right|=2-0=2,\quad A_{32}=-\left|\begin{array}{cc}1&2\\3&-2\end{array}\right|=-(-2-6)=8$
$A_{33}=\left|\begin{array}{cc}1&-1\\3&0\end{array}\right|=0+3=3$

$\text{adj}(A) = \begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}$

Step 3: Compute $A(\text{adj}A)$ :
$\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}$
$= \begin{bmatrix}0+11+0&3-1-2&2-8+6\\0+0+0&9+0+2&6+0-6\\0+0+0&3+0-3&2+0+9\end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} = 11I\checkmark$

Step 4: Compute $(\text{adj}A)A$ :
$\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}$
$= \begin{bmatrix}0+9+2&0+0+0&0-6+6\\-11+3+8&11+0+0&-22-2+24\\0-3+3&0+0+0&0+2+9\end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} = 11I\checkmark$

Hence verified.

5Find the inverse of

\begin{bmatrix}2&-2\\4&3\end{bmatrix}

(if it exists).Show solution

Given: $A = \begin{bmatrix}2&-2\\4&3\end{bmatrix}$

Step 1: $|A| = (2)(3)-(-2)(4) = 6+8 = 14 \neq 0$ , so inverse exists.

Step 2: $\text{adj}(A) = \begin{bmatrix}3&2\\-4&2\end{bmatrix}$

Step 3:
$A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}$

Answer: $A^{-1} = \dfrac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}$

6Find the inverse of

\begin{bmatrix}-1&5\\-3&2\end{bmatrix}

(if it exists).Show solution

Given: $A = \begin{bmatrix}-1&5\\-3&2\end{bmatrix}$

Step 1: $|A| = (-1)(2)-(5)(-3) = -2+15 = 13 \neq 0$

Step 2: $\text{adj}(A) = \begin{bmatrix}2&-5\\3&-1\end{bmatrix}$

Step 3:
$A^{-1} = \frac{1}{13}\begin{bmatrix}2&-5\\3&-1\end{bmatrix}$

7Find the inverse of

\begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix}

(if it exists).Show solution

Given: $A = \begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix}$ (upper triangular)

Step 1: $|A| = 1\cdot2\cdot5 = 10 \neq 0$

Step 2: Cofactors:
$A_{11}=\left|\begin{array}{cc}2&4\\0&5\end{array}\right|=10,\quad A_{12}=-\left|\begin{array}{cc}0&4\\0&5\end{array}\right|=0,\quad A_{13}=\left|\begin{array}{cc}0&2\\0&0\end{array}\right|=0$
$A_{21}=-\left|\begin{array}{cc}2&3\\0&5\end{array}\right|=-10,\quad A_{22}=\left|\begin{array}{cc}1&3\\0&5\end{array}\right|=5,\quad A_{23}=-\left|\begin{array}{cc}1&2\\0&0\end{array}\right|=0$
$A_{31}=\left|\begin{array}{cc}2&3\\2&4\end{array}\right|=8-6=2,\quad A_{32}=-\left|\begin{array}{cc}1&3\\0&4\end{array}\right|=-4,\quad A_{33}=\left|\begin{array}{cc}1&2\\0&2\end{array}\right|=2$

$\text{adj}(A) = \begin{bmatrix}10&-10&2\\0&5&-4\\0&0&2\end{bmatrix}$

Step 3:
$A^{-1} = \frac{1}{10}\begin{bmatrix}10&-10&2\\0&5&-4\\0&0&2\end{bmatrix}$

8Find the inverse of

\begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}

(if it exists).

Given: $A = \begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}$

Step 1: $|A| = 1(3\cdot(-1)-0\cdot2)-0+0 = -3 \neq 0$

Step 2: Cofactors:
$A_{11}=\left|\begin{array}{cc}3&0\\2&-1\end{array}\right|=-3,\quad A_{12}=-\left|\begin{array}{cc}3&0\\5&-1\end{array}\right|=-(-3)=3,\quad A_{13}=\left|\begin{array}{cc}3&3\\5&2\end{array}\right|=6-15=-9$
$A_{21}=-\left|\begin{array}{cc}0&0\\2&-1\end{array}\right|=0,\quad A_{22}=\left|\begin{array}{cc}1&0\\5&-1\end{array}\right|=-1,\quad A_{23}=-\left|\begin{array}{cc}1&0\\5&2\end{array}\right|=-2$
$A_{31}=\left|\begin{array}{cc}0&0\\3&0\end{array}\right|=0,\quad A_{32}=-\left|\begin{array}{cc}1&0\\3&0\end{array}\right|=0,\quad A_{33}=\left|\begin{array}{cc}1&0\\3&3\end{array}\right|=3$

$\text{adj}(A) = \begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix}$

Step 3:
$A^{-1} = \frac{1}{-3}\begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix} = \begin{bmatrix}1&0&0\\-1&\frac{1}{3}&0\\3&\frac{2}{3}&-1\end{bmatrix}$

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9Find the inverse of

\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}

(if it exists).

Given: $A = \begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 2[(-1)(1)-(0)(2)] - 1[(4)(1)-(0)(-7)] + 3[(4)(2)-(-1)(-7)]$
$= 2(-1) - 1(4) + 3(8-7)$
$= -2 - 4 + 3 = -3$

Step 2: Cofactors:
$A_{11}=(-1-0)=-1,\quad A_{12}=-(4-0)=-4,\quad A_{13}=(8-7)=1$
$A_{21}=-(1-6)=5,\quad A_{22}=(2+21)=23,\quad A_{23}=-(4+7)=-11$
$A_{31}=(0+3)=3,\quad A_{32}=-(0-12)=12,\quad A_{33}=(-2-4)=-6$

$\text{adj}(A) = \begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}$

Step 3:
$A^{-1} = \frac{1}{-3}\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix} = \begin{bmatrix}\frac{1}{3}&-\frac{5}{3}&-1\\\frac{4}{3}&-\frac{23}{3}&-4\\-\frac{1}{3}&\frac{11}{3}&2\end{bmatrix}$

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10Find the inverse of

\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}

(if it exists).

Given: $A = \begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 1(8-6)-(-1)(0+9)+2(0-6)$
$= 1(2)+1(9)+2(-6) = 2+9-12 = -1$

Step 2: Cofactors:
$A_{11}=(8-6)=2,\quad A_{12}=-(0+9)=-9,\quad A_{13}=(0-6)=-6$
$A_{21}=-(-4+4)=0,\quad A_{22}=(4-6)=-2,\quad A_{23}=-(−2+3)=-1$
$A_{31}=(3-4)=-1,\quad A_{32}=-(-3-0)=3,\quad A_{33}=(2-0)=2$

$\text{adj}(A) = \begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}$

Step 3:
$A^{-1} = \frac{1}{-1}\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix} = \begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$

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11Find the inverse of

\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}

(if it exists).

Given: $A = \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$

Step 1: $|A|$ expanding along $C_1$ :
$= 1\cdot\left|\begin{array}{cc}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{array}\right| = (-\cos^2\alpha - \sin^2\alpha) = -1$

Since $|A| = -1 \neq 0$ , inverse exists.

Step 2: Cofactors:
$A_{11}=(-\cos^2\alpha-\sin^2\alpha)=-1,\quad A_{12}=0,\quad A_{13}=0$
$A_{21}=-(0-0)=0,\quad A_{22}=(-\cos\alpha-0)=-\cos\alpha,\quad A_{23}=-(\sin\alpha-0)=-\sin\alpha$
$A_{31}=(0-0)=0,\quad A_{32}=-(\sin\alpha-0)=-\sin\alpha,\quad A_{33}=(\cos\alpha-0)=\cos\alpha$

$\text{adj}(A) = \begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}$

Step 3:
$A^{-1} = \frac{1}{-1}\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix} = \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$

(Note: $A^{-1} = A$ in this case.)

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12Let

A = \begin{bmatrix}3&7\\2&5\end{bmatrix}

and

B = \begin{bmatrix}6&8\\7&9\end{bmatrix}

. Verify that

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

Step 1: Find $AB$ :
$AB = \begin{bmatrix}3&7\\2&5\end{bmatrix}\begin{bmatrix}6&8\\7&9\end{bmatrix} = \begin{bmatrix}18+49&24+63\\12+35&16+45\end{bmatrix} = \begin{bmatrix}67&87\\47&61\end{bmatrix}$

Step 2: Find $(AB)^{-1}$ :
$|AB| = 67\times61 - 87\times47 = 4087 - 4089 = -2$
$\text{adj}(AB) = \begin{bmatrix}61&-87\\-47&67\end{bmatrix}$
$(AB)^{-1} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix}$

Step 3: Find $A^{-1}$ :
$|A| = 15-14=1,\quad A^{-1} = \begin{bmatrix}5&-7\\-2&3\end{bmatrix}$

Step 4: Find $B^{-1}$ :
$|B| = 54-56=-2,\quad B^{-1} = \frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}$

Step 5: Find $B^{-1}A^{-1}$ :
$B^{-1}A^{-1} = \frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}$
$= \frac{1}{-2}\begin{bmatrix}45+16&-63-24\\-35-12&49+18\end{bmatrix} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix}$

Since $(AB)^{-1} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} = B^{-1}A^{-1}$ , the result is verified. $\checkmark$

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13If

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

, show that

A^2 - 5A + 7I = O

. Hence find

A^{-1}

Step 1: Compute $A^2$ :
$A^2 = \begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix} = \begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix} = \begin{bmatrix}8&5\\-5&3\end{bmatrix}$

Step 2: Compute $A^2 - 5A + 7I$ :
$= \begin{bmatrix}8&5\\-5&3\end{bmatrix} - \begin{bmatrix}15&5\\-5&10\end{bmatrix} + \begin{bmatrix}7&0\\0&7\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix} = O \quad\checkmark$

Step 3: Find $A^{-1}$ :

From $A^2 - 5A + 7I = O$ :
$A^2 - 5A = -7I$
$A(A - 5I) = -7I$

Multiplying both sides by $A^{-1}$ on the right:
$A - 5I = -7A^{-1}$
$A^{-1} = \frac{1}{7}(5I - A) = \frac{1}{7}\left(\begin{bmatrix}5&0\\0&5\end{bmatrix} - \begin{bmatrix}3&1\\-1&2\end{bmatrix}\right) = \frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}$

Answer: $A^{-1} = \dfrac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}$

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14For the matrix

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

, find the numbers

a

and

b

such that

A^2 + aA + bI = O

Step 1: Compute $A^2$ :
$A^2 = \begin{bmatrix}3&2\\1&1\end{bmatrix}\begin{bmatrix}3&2\\1&1\end{bmatrix} = \begin{bmatrix}9+2&6+2\\3+1&2+1\end{bmatrix} = \begin{bmatrix}11&8\\4&3\end{bmatrix}$

Step 2: Set up $A^2 + aA + bI = O$ :
$\begin{bmatrix}11&8\\4&3\end{bmatrix} + a\begin{bmatrix}3&2\\1&1\end{bmatrix} + b\begin{bmatrix}1&0\\0&1\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$

Step 3: Equating elements:

$(1,1)$ : $11 + 3a + b = 0$
$(1,2)$ : $8 + 2a = 0 \Rightarrow a = -4$
$(2,1)$ : $4 + a = 0 \Rightarrow a = -4$ ✓
$(2,2)$ : $3 + a + b = 0 \Rightarrow 3 - 4 + b = 0 \Rightarrow b = 1$

Verification with (1,1): $11 + 3(-4) + 1 = 11 - 12 + 1 = 0$ ✓

Answer: $a = -4,\ b = 1$

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15For the matrix

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

, show that

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

. Hence find

A^{-1}

Step 1: Compute $A^2$ :
$A^2 = A\cdot A = \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}$
$= \begin{bmatrix}1+1+2&1+2-1&1-3+3\\1+2-6&1+4+3&1-6-9\\2-1+6&2-2-3&2+3+9\end{bmatrix} = \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}$

Step 2: Compute $A^3 = A^2\cdot A$ :
$= \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}$
$= \begin{bmatrix}4+2+2&4+4-1&4-6+3\\-3+8-28&-3+16+14&-3-24-42\\7-3+28&7-6-14&7+9+42\end{bmatrix} = \begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}$

Step 3: Compute $A^3 - 6A^2 + 5A + 11I$ :
$= \begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix} - 6\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix} + 5\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} + 11\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$= \begin{bmatrix}8-24+5+11&7-12+5+0&1-6+5+0\\-23+18+5+0&27-48+10+11&-69+84-15+0\\32-42+10+0&-13+18-5+0&58-84+15+11\end{bmatrix}$
$= \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} = O \quad\checkmark$

Step 4: Find $A^{-1}$ :

From $A^3 - 6A^2 + 5A + 11I = O$ , multiply by $A^{-1}$ :
$A^2 - 6A + 5I + 11A^{-1} = O$
$A^{-1} = \frac{1}{11}(6A - A^2 - 5I)$
$= \frac{1}{11}\left(6\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} - \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix} - \begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}\right)$
$= \frac{1}{11}\begin{bmatrix}6-4-5&6-2-0&6-1-0\\6+3-0&12-8-5&-18+14-0\\12-7-0&-6+3-0&18-14-5\end{bmatrix}$
$= \frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}$

Answer: $A^{-1} = \dfrac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}$

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16If

A = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}

, verify that

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

and hence find

A^{-1}

Step 1: Compute $A^2$ :
$A^2 = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}^2 = \begin{bmatrix}4+1+1&-2-2-1&2+1+2\\-2-2-1&1+4+1&-1-2-2\\2+1+2&-1-2-2&1+1+4\end{bmatrix} = \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}$

Step 2: Compute $A^3 = A^2\cdot A$ :
$= \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
$= \begin{bmatrix}12+5+5&-6-10-5&6+5+10\\-10-6-5&5+12+5&-5-6-10\\10+5+6&-5-10-6&5+5+12\end{bmatrix} = \begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}$

Step 3: Verify $A^3 - 6A^2 + 9A - 4I = O$ :
$= \begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix} - 6\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix} + 9\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix} - 4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$= \begin{bmatrix}22-36+18-4&-21+30-9-0&21-30+9-0\\-21+30-9-0&22-36+18-4&-21+30-9-0\\21-30+9-0&-21+30-9-0&22-36+18-4\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} = O\checkmark$

Step 4: Find $A^{-1}$ :

From $A^3 - 6A^2 + 9A - 4I = O$ , multiply by $A^{-1}$ :
$A^2 - 6A + 9I - 4A^{-1} = O$
$A^{-1} = \frac{1}{4}(A^2 - 6A + 9I)$
$= \frac{1}{4}\left(\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix} - \begin{bmatrix}12&-6&6\\-6&12&-6\\6&-6&12\end{bmatrix} + \begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}\right)$
$= \frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$

Answer: $A^{-1} = \dfrac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$

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17Let

A

be a nonsingular square matrix of order

3\times3

. Then

|\text{adj}A|

is equal to: (A)

|A|

, (B)

|A|^2

, (C)

|A|^3

, (D)

3|A|

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18If

A

is an invertible matrix of order 2, then

\det(A^{-1})

is equal to: (A)

\det(A)

, (B)

\frac{1}{\det(A)}

, (C) 1, (D) 0

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Exercise 4.5

1Examine the consistency of the system:

x + 2y = 2

2x + 3y = 3

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2Examine the consistency of the system:

2x - y = 5

x + y = 4

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3Examine the consistency of the system:

x + 3y = 5

2x + 6y = 8

Matrix form: $A = \begin{bmatrix}1&3\\2&6\end{bmatrix}$ , $B = \begin{bmatrix}5\\8\end{bmatrix}$

Step 1: $|A| = 6 - 6 = 0$

Step 2: Find $\text{adj}(A)$ :
$\text{adj}(A) = \begin{bmatrix}6&-3\\-2&1\end{bmatrix}$

Step 3: Compute $(\text{adj}A)B$ :
$\begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix} = \begin{bmatrix}30-24\\-10+8\end{bmatrix} = \begin{bmatrix}6\\-2\end{bmatrix} \neq \begin{bmatrix}0\\0\end{bmatrix}$

Since $|A|=0$ and $(\text{adj}A)B \neq O$ , the system is inconsistent (no solution).

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4Examine the consistency of the system:

x+y+z=1

2x+3y+2z=2

ax+ay+2az=4

Matrix form: $A = \begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}$ , $B = \begin{bmatrix}1\\2\\4\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 1(6a-2a) - 1(4a-2a) + 1(2a-3a)$
$= 4a - 2a - a = a$

Case 1: If $a \neq 0$ , then $|A| \neq 0$ , so the system is consistent (unique solution).

Case 2: If $a = 0$ , then $A = \begin{bmatrix}1&1&1\\2&3&2\\0&0&0\end{bmatrix}$ , $|A|=0$ .

$\text{adj}(A)B$ needs to be checked. With $a=0$ , the third equation becomes $0=4$ , which is impossible. So the system is inconsistent when $a=0$ .

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5Examine the consistency of the system:

3x-y-2z=2

2y-z=-1

3x-5y=3

Matrix form: $A = \begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}$ , $B = \begin{bmatrix}2\\-1\\3\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 3(0-5)-(-1)(0+3)+(-2)(0-6)$
$= 3(-5)+1(3)+(-2)(-6)$
$= -15+3+12 = 0$

Step 2: Find $(\text{adj}A)B$ . Cofactors of $A$ :
$A_{11}=(0-5)=-5,\quad A_{12}=-(0+3)=-3,\quad A_{13}=(0-6)=-6$
$A_{21}=-(0-10)=10,\quad A_{22}=(0+6)=6,\quad A_{23}=-(−15+3)=12$
$A_{31}=(1+4)=5,\quad A_{32}=-(−3-0)=3,\quad A_{33}=(6-0)=6$

$\text{adj}(A) = \begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}$

$(\text{adj}A)B = \begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}\begin{bmatrix}2\\-1\\3\end{bmatrix} = \begin{bmatrix}-10-10+15\\-6-6+9\\-12-12+18\end{bmatrix} = \begin{bmatrix}-5\\-3\\-6\end{bmatrix} \neq O$

Since $|A|=0$ and $(\text{adj}A)B \neq O$ , the system is inconsistent.

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6Examine the consistency of the system:

5x-y+4z=5

2x+3y+5z=2

5x-2y+6z=-1

Matrix form: $A = \begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}$ , $B = \begin{bmatrix}5\\2\\-1\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 5(18+10)-(-1)(12-25)+4(-4-15)$
$= 5(28)+1(-13)+4(-19)$
$= 140 - 13 - 76 = 51$

Wait, let me recompute:
$= 5(3\cdot6-5\cdot(-2))-(-1)(2\cdot6-5\cdot5)+4(2\cdot(-2)-3\cdot5)$
$= 5(18+10)+1(12-25)+4(-4-15)$
$= 5(28)+1(-13)+4(-19)$
$= 140 - 13 - 76 = 51$

Hmm, let me redo:
$|A| = 5(18-(-10))-(-1)(12-25)+4(-4-15)$
$= 5(28)+1(-13)+4(-19) = 140-13-76 = 51$

Actually: $3\times6 = 18$ , $5\times(-2)=-10$ , so $18-(-10)=28$ . ✓
$2\times6=12$ , $5\times5=25$ , so $12-25=-13$ . ✓
$2\times(-2)=-4$ , $3\times5=15$ , so $-4-15=-19$ . ✓

$|A| = 140 - 13 - 76 = 51 \neq 0$

Since $|A| \neq 0$ , the system is consistent (has a unique solution).

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7Solve using matrix method:

5x+2y=4

7x+3y=5

Matrix form: $AX = B$ , where $A = \begin{bmatrix}5&2\\7&3\end{bmatrix}$ , $X = \begin{bmatrix}x\\y\end{bmatrix}$ , $B = \begin{bmatrix}4\\5\end{bmatrix}$

Step 1: $|A| = 15-14 = 1 \neq 0$

Step 2: $A^{-1} = \frac{1}{1}\begin{bmatrix}3&-2\\-7&5\end{bmatrix} = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix} = \begin{bmatrix}12-10\\-28+25\end{bmatrix} = \begin{bmatrix}2\\-3\end{bmatrix}$

Answer: $x = 2,\ y = -3$

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8Solve using matrix method:

2x-y=-2

3x+4y=3

Matrix form: $A = \begin{bmatrix}2&-1\\3&4\end{bmatrix}$ , $B = \begin{bmatrix}-2\\3\end{bmatrix}$

Step 1: $|A| = 8+3 = 11$

Step 2: $A^{-1} = \frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$\begin{bmatrix}x\\y\end{bmatrix} = \frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix} = \frac{1}{11}\begin{bmatrix}-8+3\\6+6\end{bmatrix} = \frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix}$

Answer: $x = -\dfrac{5}{11},\ y = \dfrac{12}{11}$

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9Solve using matrix method:

4x-3y=3

3x-5y=7

Matrix form: $A = \begin{bmatrix}4&-3\\3&-5\end{bmatrix}$ , $B = \begin{bmatrix}3\\7\end{bmatrix}$

Step 1: $|A| = -20+9 = -11$

Step 2: $A^{-1} = \frac{1}{-11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$\begin{bmatrix}x\\y\end{bmatrix} = \frac{1}{-11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix} = \frac{1}{-11}\begin{bmatrix}-15+21\\-9+28\end{bmatrix} = \frac{1}{-11}\begin{bmatrix}6\\19\end{bmatrix}$

Answer: $x = -\dfrac{6}{11},\ y = -\dfrac{19}{11}$

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10Solve using matrix method:

5x+2y=3

3x+2y=5

Matrix form: $A = \begin{bmatrix}5&2\\3&2\end{bmatrix}$ , $B = \begin{bmatrix}3\\5\end{bmatrix}$

Step 1: $|A| = 10-6 = 4$

Step 2: $A^{-1} = \frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$\begin{bmatrix}x\\y\end{bmatrix} = \frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix} = \frac{1}{4}\begin{bmatrix}6-10\\-9+25\end{bmatrix} = \frac{1}{4}\begin{bmatrix}-4\\16\end{bmatrix} = \begin{bmatrix}-1\\4\end{bmatrix}$

Answer: $x = -1,\ y = 4$

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11Solve using matrix method:

2x+y+z=1

x-2y-z=\frac{3}{2}

3y-5z=9

Matrix form: $A = \begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}$ , $B = \begin{bmatrix}1\\3/2\\9\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 2(10+3) - 1(-5-0) + 1(3-0)$
$= 2(13)+5+3 = 26+5+3 = 34$

Step 2: Cofactors:
$A_{11}=(10+3)=13,\quad A_{12}=-(-5-0)=5,\quad A_{13}=(3-0)=3$
$A_{21}=-(−5-3)=8,\quad A_{22}=(-10-0)=-10,\quad A_{23}=-(6-0)=-6$
$A_{31}=(-1+2)=1,\quad A_{32}=-(-2-1)=3,\quad A_{33}=(-4-1)=-5$

$A^{-1} = \frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$\begin{bmatrix}x\\y\\z\end{bmatrix} = \frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}\begin{bmatrix}1\\3/2\\9\end{bmatrix}$
$= \frac{1}{34}\begin{bmatrix}13+12+9\\5-15+27\\3-9-45\end{bmatrix} = \frac{1}{34}\begin{bmatrix}34\\17\\-51\end{bmatrix} = \begin{bmatrix}1\\1/2\\-3/2\end{bmatrix}$

Answer: $x = 1,\ y = \dfrac{1}{2},\ z = -\dfrac{3}{2}$

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12Solve using matrix method:

x-y+z=4

2x+y-3z=0

x+y+z=2

Matrix form: $A = \begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}$ , $B = \begin{bmatrix}4\\0\\2\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 1(1+3)-(-1)(2+3)+1(2-1)$
$= 4+5+1 = 10$

Step 2: Cofactors:
$A_{11}=4,\quad A_{12}=-(2+3)=-5,\quad A_{13}=(2-1)=1$
$A_{21}=-(-1-1)=2,\quad A_{22}=(1-1)=0,\quad A_{23}=-(1+1)=-2$
$A_{31}=(3-1)=2,\quad A_{32}=-(-3-2)=5,\quad A_{33}=(1+2)=3$

$A^{-1} = \frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$= \frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\begin{bmatrix}4\\0\\2\end{bmatrix} = \frac{1}{10}\begin{bmatrix}16+0+4\\-20+0+10\\4+0+6\end{bmatrix} = \frac{1}{10}\begin{bmatrix}20\\-10\\10\end{bmatrix} = \begin{bmatrix}2\\-1\\1\end{bmatrix}$

Answer: $x = 2,\ y = -1,\ z = 1$

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13Solve using matrix method:

2x+3y+3z=5

x-2y+z=-4

3x-y-2z=3

Matrix form: $A = \begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}$ , $B = \begin{bmatrix}5\\-4\\3\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 2(4+1)-3(-2-3)+3(-1+6)$
$= 2(5)-3(-5)+3(5) = 10+15+15 = 40$

Step 2: Cofactors:
$A_{11}=(4+1)=5,\quad A_{12}=-(-2-3)=5,\quad A_{13}=(-1+6)=5$
$A_{21}=-(−6+3)=3,\quad A_{22}=(-4-9)=-13,\quad A_{23}=-(-2-9)=11$
$A_{31}=(3+6)=9,\quad A_{32}=-(2-3)=1,\quad A_{33}=(-4-3)=-7$

$A^{-1} = \frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$= \frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix} = \frac{1}{40}\begin{bmatrix}25-12+27\\25+52+3\\25-44-21\end{bmatrix} = \frac{1}{40}\begin{bmatrix}40\\80\\-40\end{bmatrix} = \begin{bmatrix}1\\2\\-1\end{bmatrix}$

Answer: $x = 1,\ y = 2,\ z = -1$

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14Solve using matrix method:

x-y+2z=7

3x+4y-5z=-5

2x-y+3z=12

Matrix form: $A = \begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}$ , $B = \begin{bmatrix}7\\-5\\12\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 1(12-5)-(-1)(9+10)+2(-3-8)$
$= 7+19-22 = 4$

Step 2: Cofactors:
$A_{11}=(12-5)=7,\quad A_{12}=-(9+10)=-19,\quad A_{13}=(-3-8)=-11$
$A_{21}=-(-3+2)=1,\quad A_{22}=(3-4)=-1,\quad A_{23}=-(-1+2)=-1$
$A_{31}=(5-8)=-3,\quad A_{32}=-(-5-6)=11,\quad A_{33}=(4+3)=7$

$A^{-1} = \frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$= \frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix} = \frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix} = \frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix}$

Answer: $x = 2,\ y = 1,\ z = 3$

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15If

A = \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}

, find

A^{-1}

. Using

A^{-1}

solve:

2x-3y+5z=11

3x+2y-4z=-5

x+y-2z=-3

Step 1: Find $|A|$ :
$|A| = 2(2\cdot(-2)-(-4)\cdot1)-(-3)(3\cdot(-2)-(-4)\cdot1)+5(3\cdot1-2\cdot1)$
$= 2(-4+4)+3(-6+4)+5(3-2)$
$= 0+3(-2)+5(1) = 0-6+5 = -1$

Step 2: Cofactors:
$A_{11}=(-4+4)=0,\quad A_{12}=-(-6+4)=2,\quad A_{13}=(3-2)=1$
$A_{21}=-(6-5)=-1,\quad A_{22}=(-4-5)=-9,\quad A_{23}=-(2+3)=-5$
$A_{31}=(12-10)=2,\quad A_{32}=-(-8-15)=23,\quad A_{33}=(4+9)=13$

$A^{-1} = \frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix} = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}$

Step 3: The system is $AX = B$ where $B = \begin{bmatrix}11\\-5\\-3\end{bmatrix}$ .

$X = A^{-1}B = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$= \begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix} = \begin{bmatrix}1\\2\\3\end{bmatrix}$

Answer: $x = 1,\ y = 2,\ z = 3$

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16The 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.

Let cost per kg of onion = $x$ , wheat = $y$ , rice = $z$ (in ₹).

System of equations:
$4x + 3y + 2z = 60$
$2x + 4y + 6z = 90$
$6x + 2y + 3z = 70$

Matrix form: $AX = B$ , where $A = \begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}$ , $B = \begin{bmatrix}60\\90\\70\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 4(12-12)-3(6-36)+2(4-24)$
$= 4(0)-3(-30)+2(-20)$
$= 0+90-40 = 50$

Step 2: Cofactors:
$A_{11}=(12-12)=0,\quad A_{12}=-(6-36)=30,\quad A_{13}=(4-24)=-20$
$A_{21}=-(9-4)=-5,\quad A_{22}=(12-12)=0,\quad A_{23}=-(8-18)=10$
$A_{31}=(18-8)=10,\quad A_{32}=-(24-4)=-20,\quad A_{33}=(16-6)=10$

$A^{-1} = \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$= \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}$
$= \frac{1}{50}\begin{bmatrix}0-450+700\\1800+0-1400\\-1200+900+700\end{bmatrix} = \frac{1}{50}\begin{bmatrix}250\\400\\400\end{bmatrix} = \begin{bmatrix}5\\8\\8\end{bmatrix}$

Answer: Cost of onion = ₹5/kg, wheat = ₹8/kg, rice = ₹8/kg.

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Miscellaneous Exercises on Chapter 4

1Prove that the determinant

\left|\begin{array}{ccc}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{array}\right|

is independent of

\theta

Expanding along $R_1$ :
$\Delta = x\left|\begin{array}{cc}-x&1\\1&x\end{array}\right| - \sin\theta\left|\begin{array}{cc}-\sin\theta&1\\\cos\theta&x\end{array}\right| + \cos\theta\left|\begin{array}{cc}-\sin\theta&-x\\\cos\theta&1\end{array}\right|$

$= x(-x^2-1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$

$= -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \cos\theta\sin\theta + x\cos^2\theta$

$= -x^3 - x + x(\sin^2\theta + \cos^2\theta)$

$= -x^3 - x + x = -x^3$

Since $\Delta = -x^3$ , which does not contain $\theta$ , the determinant is independent of $\theta$ . $\blacksquare$

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2Evaluate

\left|\begin{array}{ccc}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{array}\right|

Expanding along $R_2$ (or $C_3$ ):

Expanding along $C_3$ :
$\Delta = -\sin\alpha\left|\begin{array}{cc}-\sin\beta&\cos\beta\\\sin\alpha\sin\beta&\sin\alpha\cos\beta\end{array}\right|\cdot(-1)^{1+3} + 0 + \cos\alpha\left|\begin{array}{cc}\cos\alpha\cos\beta&\cos\alpha\sin\beta\\-\sin\beta&\cos\beta\end{array}\right|\cdot(-1)^{3+3}$

Wait, let me expand along $C_3$ :
$\Delta = (-\sin\alpha)\cdot(-1)^{1+3}\left|\begin{array}{cc}-\sin\beta&\cos\beta\\\sin\alpha\sin\beta&\sin\alpha\cos\beta\end{array}\right| + 0\cdot(\ldots) + \cos\alpha\cdot(-1)^{3+3}\left|\begin{array}{cc}\cos\alpha\cos\beta&\cos\alpha\sin\beta\\-\sin\beta&\cos\beta\end{array}\right|$

$= (-\sin\alpha)(1)(-\sin\beta\cdot\sin\alpha\cos\beta - \cos\beta\cdot\sin\alpha\sin\beta) + \cos\alpha(1)(\cos\alpha\cos\beta\cdot\cos\beta - \cos\alpha\sin\beta\cdot(-\sin\beta))$

$= (-\sin\alpha)(-\sin\alpha\sin\beta\cos\beta - \sin\alpha\cos\beta\sin\beta) + \cos\alpha(\cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta)$

$= (-\sin\alpha)(-2\sin\alpha\sin\beta\cos\beta) + \cos\alpha(\cos\alpha)$

Hmm, let me redo more carefully.

$M_{13} = \left|\begin{array}{cc}-\sin\beta&\cos\beta\\\sin\alpha\sin\beta&\sin\alpha\cos\beta\end{array}\right| = -\sin\beta\cdot\sin\alpha\cos\beta - \cos\beta\cdot\sin\alpha\sin\beta = -\sin\alpha\sin\beta\cos\beta - \sin\alpha\sin\beta\cos\beta = -2\sin\alpha\sin\beta\cos\beta$

Wait: $(-\sin\beta)(\sin\alpha\cos\beta) - (\cos\beta)(\sin\alpha\sin\beta) = -\sin\alpha\sin\beta\cos\beta - \sin\alpha\cos\beta\sin\beta = -2\sin\alpha\sin\beta\cos\beta$

$M_{33} = \left|\begin{array}{cc}\cos\alpha\cos\beta&\cos\alpha\sin\beta\\-\sin\beta&\cos\beta\end{array}\right| = \cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta = \cos\alpha$

So:
$\Delta = (-\sin\alpha)(+1)(-2\sin\alpha\sin\beta\cos\beta) + \cos\alpha(+1)(\cos\alpha)$
$= 2\sin^2\alpha\sin\beta\cos\beta + \cos^2\alpha$
$= \sin^2\alpha\sin2\beta + \cos^2\alpha$

This doesn't simplify to a clean answer. Let me try expanding along $R_2$ instead.

Expanding along $R_2$ :
$\Delta = -(-\sin\beta)\left|\begin{array}{cc}\cos\alpha\sin\beta&-\sin\alpha\\\sin\alpha\sin\beta&\cos\alpha\end{array}\right| + \cos\beta\left|\begin{array}{cc}\cos\alpha\cos\beta&-\sin\alpha\\\sin\alpha\cos\beta&\cos\alpha\end{array}\right| - 0$

$= \sin\beta(\cos\alpha\sin\beta\cos\alpha + \sin\alpha\cdot\sin\alpha\sin\beta) + \cos\beta(\cos\alpha\cos\beta\cos\alpha + \sin\alpha\sin\alpha\cos\beta)$

$= \sin\beta\cdot\sin\beta(\cos^2\alpha + \sin^2\alpha) + \cos\beta\cdot\cos\beta(\cos^2\alpha + \sin^2\alpha)$

$= \sin^2\beta(1) + \cos^2\beta(1) = 1$

Answer: $\Delta = 1$

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3If

A^{-1} = \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}

and

B = \begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}

, find

(AB)^{-1}

Using the property: $(AB)^{-1} = B^{-1}A^{-1}$

Step 1: Find $B^{-1}$ .

$|B|$ expanding along $R_1$ :
$= 1(3-0)-2(-1-0)+(-2)(2-0) = 3+2-4 = 1$

Cofactors of $B$ :
$B_{11}=(3-0)=3,\quad B_{12}=-(-1-0)=1,\quad B_{13}=(2-0)=2$
$B_{21}=-(2-4)=2,\quad B_{22}=(1-0)=1,\quad B_{23}=-(-2-0)=2$
$B_{31}=(0+6)=6,\quad B_{32}=-(0-2)=2,\quad B_{33}=(3-2)=5$

$B^{-1} = \frac{1}{1}\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}$

Step 2: $(AB)^{-1} = B^{-1}A^{-1}$ :
$= \begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
$= \begin{bmatrix}9-30+30&-3+12-12&3-10+12\\3-15+10&-1+6-4&1-5+4\\6-30+25&-2+12-10&2-10+10\end{bmatrix}$
$= \begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}$

Answer: $(AB)^{-1} = \begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}$

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4Let

A = \begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}

. Verify that (i)

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

, (ii)

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

Step 1: Find $|A|$ :
$|A| = 1(15-1)-2(10-1)+1(2-3) = 14-18-1 = -5$

Step 2: Find cofactors of $A$ :
$A_{11}=14,\quad A_{12}=-9,\quad A_{13}=-1$
$A_{21}=-9,\quad A_{22}=4,\quad A_{23}=1$
$A_{31}=-1,\quad A_{32}=1,\quad A_{33}=-1$

$\text{adj}(A) = \begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}$

$A^{-1} = \frac{1}{-5}\begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}$

Verification of (ii): $(A^{-1})^{-1} = A$

Since $A^{-1} = \frac{1}{|A|}\text{adj}(A)$ , we have $(A^{-1})^{-1} = A$ by the standard property of invertible matrices (inverting twice returns the original matrix). This follows from $A\cdot A^{-1} = I \Rightarrow (A^{-1})^{-1} = A$ . ✓

Verification of (i): $[\text{adj}A]^{-1} = \text{adj}(A^{-1})$

We know $\text{adj}(A) = |A|\cdot A^{-1}$ , so $[\text{adj}A]^{-1} = \frac{1}{|A|}A$ .

Also, $\text{adj}(A^{-1}) = |A^{-1}|\cdot(A^{-1})^{-1} = \frac{1}{|A|}\cdot A$ .

Hence $[\text{adj}A]^{-1} = \text{adj}(A^{-1})$ . ✓

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5Evaluate

\left|\begin{array}{ccc}x&y&x+y\\y&x+y&x\\x+y&x&y\end{array}\right|

Apply $C_1 \to C_1 + C_2 + C_3$ :

Each row sum: $x+y+(x+y) = 2(x+y)$

$\Delta = \left|\begin{array}{ccc}2(x+y)&y&x+y\\2(x+y)&x+y&x\\2(x+y)&x&y\end{array}\right|$

Take $2(x+y)$ common from $C_1$ :
$= 2(x+y)\left|\begin{array}{ccc}1&y&x+y\\1&x+y&x\\1&x&y\end{array}\right|$

Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ :
$= 2(x+y)\left|\begin{array}{ccc}1&y&x+y\\0&x&-y\\0&x-y&-x\end{array}\right|$

Expanding along $C_1$ :
$= 2(x+y)\cdot 1\cdot\left|\begin{array}{cc}x&-y\\x-y&-x\end{array}\right|$
$= 2(x+y)(-x^2 + y(x-y))$
$= 2(x+y)(-x^2 + xy - y^2)$
$= -2(x+y)(x^2 - xy + y^2)$
$= -2(x^3 + y^3)$

Answer: $\Delta = -2(x^3+y^3)$

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6Evaluate

\left|\begin{array}{ccc}1&x&y\\1&x+y&y\\1&x&x+y\end{array}\right|

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7Solve the system of equations:

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

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

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

Substitution: Let $u = \dfrac{1}{x},\ v = \dfrac{1}{y},\ w = \dfrac{1}{z}$ .

System becomes:
$2u + 3v + 10w = 4$
$4u - 6v + 5w = 1$
$6u + 9v - 20w = 2$

Matrix form: $A = \begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}$ , $B = \begin{bmatrix}4\\1\\2\end{bmatrix}$

Step 1: $|A|$ expanding along $R_1$ :
$= 2(120-45)-3(-80-30)+10(36+36)$
$= 2(75)-3(-110)+10(72)$
$= 150+330+720 = 1200$

Step 2: Cofactors:
$A_{11}=(120-45)=75,\quad A_{12}=-(-80-30)=110,\quad A_{13}=(36+36)=72$
$A_{21}=-(−60-90)=150,\quad A_{22}=(-40-60)=-100,\quad A_{23}=-(18-18)=0$
$A_{31}=(15+60)=75,\quad A_{32}=-(10-40)=30,\quad A_{33}=(-12-12)=-24$

$A^{-1} = \frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}$

Step 3: $X = A^{-1}B$ :
$\begin{bmatrix}u\\v\\w\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}$
$= \frac{1}{1200}\begin{bmatrix}300+150+150\\440-100+60\\288+0-48\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix} = \begin{bmatrix}1/2\\1/3\\1/5\end{bmatrix}$

So $u = \dfrac{1}{2},\ v = \dfrac{1}{3},\ w = \dfrac{1}{5}$ .

Answer: $x = 2,\ y = 3,\ z = 5$

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8If

x, y, z

are nonzero real numbers, then the inverse of matrix

A = \begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}

is: (A)

\begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}

, (B)

xyz\begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}

, (C)

\frac{1}{xyz}\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}

, (D)

\frac{1}{xyz}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

Correct Option: (A)

$|A| = xyz$ . The cofactor matrix of a diagonal matrix is also diagonal with entries $yz, xz, xy$ .

$\text{adj}(A) = \begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}$

$A^{-1} = \frac{1}{xyz}\begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix} = \begin{bmatrix}1/x&0&0\\0&1/y&0\\0&0&1/z\end{bmatrix} = \begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}$

Answer: (A)

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9Let

A = \begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}

, where

0\leq\theta\leq2\pi

. Then: (A)

\det(A)=0

, (B)

\det(A)\in(2,\infty)

, (C)

\det(A)\in(2,4)

, (D)

\det(A)\in[2,4]

Correct Option: (D) $\det(A)\in[2,4]$

Expanding along $R_1$ :
$\det(A) = 1(1+\sin^2\theta) - \sin\theta(-\sin\theta+\sin\theta) + 1(\sin^2\theta+1)$
$= (1+\sin^2\theta) - \sin\theta(0) + (1+\sin^2\theta)$
$= 2(1+\sin^2\theta) = 2 + 2\sin^2\theta$

Since $0 \leq \sin^2\theta \leq 1$ :
$2 \leq 2+2\sin^2\theta \leq 4$

So $\det(A) \in [2, 4]$ .

Answer: (D)

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