EXERCISE 4.1 ( Determinants )

Evaluate the determinant:(Class 12 ncert solution math exercise 4.1)

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

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

Hence,

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

$=2(-1)-4(-5)$

$=-2+20$

$=18$

Question 2: Evaluate the determinants:

(i) $\begin{vmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{vmatrix}$

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

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

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

Question 3: If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]$ , then show that $|2 A|=4|A|$

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

Therefore,

$2 A =2\left[\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right]$
$=\left[\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right]$

Hence,

$L.H.S. =|2 A|$
$=\left|\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right|$
$=2 \times 4-4 \times 8$
$=8-32$
$=-24$

Now,

$|A| =\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=1 \times 2-2 \times 4$
$=2-8$
$=-6$

Therefore,

$R.H.S. =4|A|$
$=4(-6)$
$=-24$

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

Question 4: If $A=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]$ , then show that $|3 A|=27|A|$

Solution:The given matrix is

$A=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]$

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

$|A| =1\left|\begin{array}{ll} 1 & 2 \\ 0 & 4 \end{array}\right|-0\left|\begin{array}{ll} 0 & 1 \\ 1 & 4 \end{array}\right|+0\left|\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right|$

$=1(4-0)-0+0$
$=4$

Therefore,

$27|A| =27|4|$

$=108$

Now,

$3 A=3\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]$

$=\left[\begin{array}{ccc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right]$

Therefore,

$|3 A| =3\left|\begin{array}{lc} 3 & 6 \\ 0 & 12 \end{array}\right|-0\left|\begin{array}{cc} 0 & 3 \\ 0 & 12 \end{array}\right|+0\left|\begin{array}{ll} 0 & 3 \\ 3 & 6 \end{array}\right|$

$=3(36-0)$
$=36(36)$
$=108$

From equations $(1)$ and $(2)$ ,

$|3 A|=27|A|$

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

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

Solution: $A=\left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right|$

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

Hence,

$|A| =-0\left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right|+0\left|\begin{array}{cc} 3 & -2 \\ 3 & 0 \end{array}\right|-(-1)\left|\begin{array}{ll} 3 & -1 \\ 3 & -5 \end{array}\right|$
$=(-15+3)$
$=-12$

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

Hence,

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

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

Hence,

$|A| =0\left|\begin{array}{cc} 0 & -3 \\ 3 & 0 \end{array}\right|-1\left|\begin{array}{cc} -1 & -3 \\ -2 & 0 \end{array}\right|+2\left|\begin{array}{cc} -1 & 0 \\ -2 & 3 \end{array}\right|$

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

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

$=6-6=0$

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

Hence,

$|A| =2\left|\begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array}\right|-0\left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right|+3\left|\begin{array}{cc} -1 & -2 \\ 2 & -1 \end{array}\right|$
$=2(0-5)-0+3(1+4)$
$=-10+15=5$

Question 6:If $A=\left[\begin{array}{ccc} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right]$ , find $|A|$

Solution: Let $A=\left[\begin{array}{ccc} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right]$

Hence,

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

Question 7: Find the values of $x$ , if
(i) $\left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$
(ii) $\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|$

Solution: (i) $\left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$
$\Rightarrow 2\times 1-4\times 5 = 2x\times x-6\times 4$
$\Rightarrow 2-20 = 2x^2-24$
$\Rightarrow -18 +24 =2x^2$
$\Rightarrow x^2 = 3$
$\Rightarrow x =\pm\sqrt{3}$

Therefore,

(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,

$\Rightarrow 2 \times 5-3 \times 4=x \times 5-3 \times 2 x$
$\Rightarrow 10-12=5 x-6 x$
$\Rightarrow-2=-x$
$\Rightarrow x=2$

Question 8: 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