Q: Find the values of $x$, if (i) $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$ (ii) $\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|$

(i) $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$ Therefore, $$ \begin{aligned} & \Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4 \\ & \Rightarrow 2-20=2 x^{2}-24 \\ & \Rightarrow 2 x^{2}=6 \\ & \Rightarrow x^{2}=3 \\ & \Rightarrow x= \pm \sqrt{3} \end{aligned} $$ (ii) $\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|$ Therefore, $$ \begin{aligned} & \Rightarrow 2 \times 5-3 \times 4=x \times 5-3 \times 2 x \\ & \Rightarrow 10-12=5 x-6 x \\ & \Rightarrow-2=-x \\ & \Rightarrow x=2 \end{aligned} $$

Q: If $\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|$, the $x$ is equal to (A) 6 (B) $\pm 6$ (C) -6 (D) 0

$\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|$ Therefore, $$ \begin{aligned} & \Rightarrow x^{2}-36=36-36 \\ & \Rightarrow x^{2}-36=0 \\ & \Rightarrow x^{2}=36 \\ & \Rightarrow x=+6 \end{aligned} $$ Thus, the correct option is B. \section*{EXERCISE 4.2}

Q: If $A=\left(\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right)$, find $|A|$

Let $A=\left(\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right)$ Hence, $$ \begin{aligned} |A| & =1\left|\begin{array}{ll} 1 & -3 \\ 4 & -9 \end{array}\right|-1\left|\begin{array}{ll} 2 & -3 \\ 5 & -9 \end{array}\right|-2\left|\begin{array}{ll} 2 & 1 \\ 5 & 4 \end{array}\right| \\ & =1(-9+12)-1(-18+15)-2(8-5) \\ & =1(3)-1(-3)-2(3) \\ & =3+3-6 \\ & =0 \end{aligned} $$

Q: Evaluate the determinant $\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|$

Let $|A|=\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|$ Hence, $$ \begin{aligned} |A| & =\left|\begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array}\right| \\ & =2(-1)-4(-5) \\ & =-2+20 \\ & =18 \end{aligned} $$

Q: Evaluate the determinants: (i) $\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|$ (ii) $\left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$

(i) $$ \begin{aligned} \left|\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right| & =(\cos \theta)(\cos \theta)-(-\sin \theta)(\sin \theta) \\ & =\cos ^{2} \theta+\sin ^{2} \theta \\ & =1 \end{aligned} $$ (ii) $$ \begin{aligned} \left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right| & =\left(x^{2}-x+1\right)(x+1)-(x-1)(x+1) \\ & =x^{3}-x^{2}+x+x^{2}-x+1-\left(x^{2}-1\right) \\ & =x^{3}+1-x^{2}+1 \\ & =x^{3}-x^{2}+2 \end{aligned} $$

Q: If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]$, then show that $|2 A|=4|A|$

The given matrix is $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]$ Therefore, $$ \begin{aligned} 2 A & =2\left(\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right) \\ & =\left(\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right) \end{aligned} $$ Hence, $$ \begin{aligned} L H S & =|2 A| \\ & =\left|\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right| \\ & =2 \times 4-4 \times 8 \\ & =8-32 \\ & =-24 \end{aligned} $$ Now, $$ \begin{aligned} |A| & =\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=1 \times 2-2 \times 4 \\ & =2-8 \\ & =-6 \end{aligned} $$ Therefore, $$ \begin{aligned} R H S & =4|A| \\ & =4(-6) \\ & =-24 \end{aligned} $$ Thus, $|2 A|=4|A|$ proved.

Q: Evaluate the determinants (i) $\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$ (ii) $\left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|$ (iii) $\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$ (iv) $\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|$

(i) Let $A=\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$ It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation. Hence, $$ \begin{aligned} |A| & =-0\left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right|+0\left|\begin{array}{cc} 3 & -2 \\ 3 & 0 \end{array}\right|-(-1)\left|\begin{array}{cc} 3 & -1 \\ 3 & -5 \end{array}\right| \\ & =(-15+3) \\ & =-12 \end{aligned} $$ (ii) Let $A=\left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|$ Hence, $$ \begin{aligned} |A| & =3\left|\begin{array}{cc} 1 & -2 \\ 3 & 1 \end{array}\right|+4\left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right|+5\left|\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array}\right| \\ & =3(1+6)+4(1+4)+5(3-2) \\ & =3(7)+4(5)+5(1) \\ & =21+20+5 \\ & =46 \end{aligned} $$ (iii) Let $A=\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$ Hence, $$ \begin{aligned} |A| & =0\left|\begin{array}{cc} 0 & -3 \\ 3 & 0 \end{array}\right|-1\left|\begin{array}{cc} -1 & -3 \\ -2 & 0 \end{array}\right|+2\left|\begin{array}{cc} -1 & 0 \\ -2 & 3 \end{array}\right| \\ & =0-1(0-6)+2(-3-0) \\ & =-1(-6)+2(-3) \\ & =6-6=0 \end{aligned} $$ (iv) Let $A=\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|$ Hence, $$ \begin{aligned} |A| & =2\left|\begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array}\right|-0\left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right|+3\left|\begin{array}{cc} -1 & -2 \\ 2 & -1 \end{array}\right| \\ & =2(0-5)-0+3(1+4) \\ & =-10+15=5 \end{aligned} $$

Question 1

Find the values of $x$, if

(i) $\left|\begin{array}{ll}2 & 4 \ 5 & 1\end{array}ight|=\left|\begin{array}{cc}2 x & 4 \ 6 & x\end{array}ight|$

(ii) $\left|\begin{array}{ll}2 & 3 \ 4 & 5\end{array}ight|=\left|\begin{array}{cc}x & 3 \ 2 x & 5\end{array}ight|$

Accepted Answer

(i) $\left|\begin{array}{ll}2 & 4 \ 5 & 1\end{array}ight|=\left|\begin{array}{cc}2 x & 4 \ 6 & x\end{array}ight|$

Therefore,

$$

\begin{aligned}

& \Rightarrow 2 	imes 1-5 	imes 4=2 x 	imes x-6 	imes 4 \

& \Rightarrow 2-20=2 x^{2}-24 \

& \Rightarrow 2 x^{2}=6 \

& \Rightarrow x^{2}=3 \

& \Rightarrow x= \pm \sqrt{3}

\end{aligned}

$$

(ii) $\left|\begin{array}{ll}2 & 3 \ 4 & 5\end{array}ight|=\left|\begin{array}{cc}x & 3 \ 2 x & 5\end{array}ight|$

Therefore,

$$

\begin{aligned}

& \Rightarrow 2 	imes 5-3 	imes 4=x 	imes 5-3 	imes 2 x \

& \Rightarrow 10-12=5 x-6 x \

& \Rightarrow-2=-x \

& \Rightarrow x=2

\end{aligned}

$$

Question 2

If $\left|\begin{array}{cc}x & 2 \ 18 & x\end{array}ight|=\left|\begin{array}{cc}6 & 2 \ 18 & 6\end{array}ight|$, the $x$ is equal to

(A) 6

(B) $\pm 6$

(C) -6

(D) 0

Accepted Answer

$\left|\begin{array}{cc}x & 2 \ 18 & x\end{array}ight|=\left|\begin{array}{cc}6 & 2 \ 18 & 6\end{array}ight|$

Therefore,

$$

\begin{aligned}

& \Rightarrow x^{2}-36=36-36 \

& \Rightarrow x^{2}-36=0 \

& \Rightarrow x^{2}=36 \

& \Rightarrow x=+6

\end{aligned}

$$

Thus, the correct option is B.

\section*{EXERCISE 4.2}

Question 3

If $A=\left(\begin{array}{lll}1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9\end{array}ight)$, find $|A|$

Accepted Answer

Let $A=\left(\begin{array}{lll}1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9\end{array}ight)$

Hence,

$$

\begin{aligned}

|A| & =1\left|\begin{array}{ll}

1 & -3 \

4 & -9

\end{array}ight|-1\left|\begin{array}{ll}

2 & -3 \

5 & -9

\end{array}ight|-2\left|\begin{array}{ll}

2 & 1 \

5 & 4

\end{array}ight| \

& =1(-9+12)-1(-18+15)-2(8-5) \

& =1(3)-1(-3)-2(3) \

& =3+3-6 \

& =0

\end{aligned}

$$

Question 4

Evaluate the determinant $\left|\begin{array}{cc}2 & 4 \ -5 & -1\end{array}ight|$

Accepted Answer

Let $|A|=\left|\begin{array}{cc}2 & 4 \ -5 & -1\end{array}ight|$

Hence,

$$

\begin{aligned}

|A| & =\left|\begin{array}{cc}

2 & 4 \

-5 & -1

\end{array}ight| \

& =2(-1)-4(-5) \

& =-2+20 \

& =18

\end{aligned}

$$

Question 5

Evaluate the determinants:

(i) $\left|\begin{array}{cc}\cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta\end{array}ight|$

(ii) $\left|\begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1\end{array}ight|$

Accepted Answer

(i)

$$

\begin{aligned}

\left|\begin{array}{cc}

\cos 	heta & -\sin 	heta \

\sin 	heta & \cos 	heta

\end{array}ight| & =(\cos 	heta)(\cos 	heta)-(-\sin 	heta)(\sin 	heta) \

& =\cos ^{2} 	heta+\sin ^{2} 	heta \

& =1

\end{aligned}

$$

(ii)

$$

\begin{aligned}

\left|\begin{array}{cc}

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

x+1 & x+1

\end{array}ight| & =\left(x^{2}-x+1ight)(x+1)-(x-1)(x+1) \

& =x^{3}-x^{2}+x+x^{2}-x+1-\left(x^{2}-1ight) \

& =x^{3}+1-x^{2}+1 \

& =x^{3}-x^{2}+2

\end{aligned}

$$

Question 6

If $A=\left[\begin{array}{ll}1 & 2 \ 4 & 2\end{array}ight]$, then show that $|2 A|=4|A|$

Accepted Answer

The given matrix is $A=\left[\begin{array}{ll}1 & 2 \ 4 & 2\end{array}ight]$

Therefore,

$$

\begin{aligned}

2 A & =2\left(\begin{array}{ll}

1 & 2 \

4 & 2

\end{array}ight) \

& =\left(\begin{array}{ll}

2 & 4 \

8 & 4

\end{array}ight)

\end{aligned}

$$

Hence,

$$

\begin{aligned}

L H S & =|2 A| \

& =\left|\begin{array}{ll}

2 & 4 \

8 & 4

\end{array}ight| \

& =2 	imes 4-4 	imes 8 \

& =8-32 \

& =-24

\end{aligned}

$$

Now,

$$

\begin{aligned}

|A| & =\left|\begin{array}{ll}

1 & 2 \

4 & 2

\end{array}ight|=1 	imes 2-2 	imes 4 \

& =2-8 \

& =-6

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

R H S & =4|A| \

& =4(-6) \

& =-24

\end{aligned}

$$

Thus, $|2 A|=4|A|$ proved.

Question 7

If $A=\left(\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4\end{array}ight)$, then show that $|3 A|=27|A|$

Accepted Answer

The given matrix is

$$

A=\left(\begin{array}{lll}

1 & 0 & 1 \

0 & 1 & 2 \

0 & 0 & 4

\end{array}ight)

$$

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column $\left(C_{1}ight)$ for easier calculation.

$$

\begin{aligned}

|A| & =1\left|\begin{array}{ll}

1 & 2 \

0 & 4

\end{array}ight|-0\left|\begin{array}{ll}

0 & 1 \

1 & 4

\end{array}ight|+0\left|\begin{array}{ll}

0 & 1 \

1 & 2

\end{array}ight| \

& =1(4-0)-0+0 \

& =4

\end{aligned}

$$

Therefore,

$$

\begin{align*}

27|A| & =27|4| \

& =108 	ag{1}

\end{align*}

$$

Now,

$$

3 A=3\left(\begin{array}{lll}

1 & 0 & 1 \

0 & 1 & 2 \

0 & 0 & 4

\end{array}ight)=\left(\begin{array}{ccc}

3 & 0 & 3 \

0 & 3 & 6 \

0 & 0 & 12

\end{array}ight)

$$

Therefore,

$$

\begin{align*}

|3 A| & =3\left|\begin{array}{cc}

3 & 6 \

0 & 12

\end{array}ight|-0\left|\begin{array}{cc}

0 & 3 \

0 & 12

\end{array}ight|+0\left|\begin{array}{cc}

0 & 3 \

3 & 6

\end{array}ight| \

& =3(36-0) \

& =36(36) \

& =108 	ag{2}

\end{align*}

$$

From equations (1) and (2),

$$

|3 A|=27|A|

$$

Thus, $|3 A|=27|A|$ proved.

Question 8

Evaluate the determinants

(i) $\left|\begin{array}{ccc}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}ight|$

(ii) $\left|\begin{array}{ccc}3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1\end{array}ight|$

(iii) $\left|\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0\end{array}ight|$

(iv) $\left|\begin{array}{ccc}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}ight|$

Accepted Answer

(i) Let $A=\left|\begin{array}{ccc}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}ight|$

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

Hence,

$$

\begin{aligned}

|A| & =-0\left|\begin{array}{cc}

-1 & -2 \

-5 & 0

\end{array}ight|+0\left|\begin{array}{cc}

3 & -2 \

3 & 0

\end{array}ight|-(-1)\left|\begin{array}{cc}

3 & -1 \

3 & -5

\end{array}ight| \

& =(-15+3) \

& =-12

\end{aligned}

$$

(ii) Let $A=\left|\begin{array}{ccc}3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1\end{array}ight|$

Hence,

$$

\begin{aligned}

|A| & =3\left|\begin{array}{cc}

1 & -2 \

3 & 1

\end{array}ight|+4\left|\begin{array}{cc}

1 & -2 \

2 & 1

\end{array}ight|+5\left|\begin{array}{cc}

1 & 1 \

2 & 3

\end{array}ight| \

& =3(1+6)+4(1+4)+5(3-2) \

& =3(7)+4(5)+5(1) \

& =21+20+5 \

& =46

\end{aligned}

$$

(iii) Let $A=\left|\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0\end{array}ight|$

Hence,

$$

\begin{aligned}

|A| & =0\left|\begin{array}{cc}

0 & -3 \

3 & 0

\end{array}ight|-1\left|\begin{array}{cc}

-1 & -3 \

-2 & 0

\end{array}ight|+2\left|\begin{array}{cc}

-1 & 0 \

-2 & 3

\end{array}ight| \

& =0-1(0-6)+2(-3-0) \

& =-1(-6)+2(-3) \

& =6-6=0

\end{aligned}

$$

(iv) Let $A=\left|\begin{array}{ccc}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}ight|$

Hence,

$$

\begin{aligned}

|A| & =2\left|\begin{array}{cc}

2 & -1 \

-5 & 0

\end{array}ight|-0\left|\begin{array}{cc}

-1 & -2 \

-5 & 0

\end{array}ight|+3\left|\begin{array}{cc}

-1 & -2 \

2 & -1

\end{array}ight| \

& =2(0-5)-0+3(1+4) \

& =-10+15=5

\end{aligned}

$$