Solution

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

Solution

(i) $ \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{vmatrix} =(\cos \theta)(\cos \theta)-(-\sin \theta)(\sin \theta)=\cos ^{2} \theta+\sin ^{2} \theta=1$

(ii) $ \begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1\end{vmatrix} $

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

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

$=x^{3}+1-x^{2}+1$

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

Solution

The given matrix is $A= \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} $.

$\therefore 2 A=2 \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} $

$\therefore$ L.H.S. $=|2 A|= \begin{vmatrix} 2 & 4 \\ 8 & 4\end{vmatrix} =2 \times 4-4 \times 8=8-32=-24$

Now, $|A|= \begin{vmatrix} 1 & 2 \\ 4 & 2\end{vmatrix} =1 \times 2-2 \times 4=2-8=-6$

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

$\therefore$ L.H.S. $=$ R.H.S.

Solution

The given matrix is $A= \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{vmatrix} $

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column $(C_1.$ ) for easier calculation.

$$ \begin{align*} & |A|=1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} -0 \begin{vmatrix} 0 & 1 \\ 0 & 4 \end{vmatrix} +0 \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} =1(4-0)-0+0=4 \\ & \therefore 27|A|=27(4)=108 \tag{i}\\ & \text{ Now, } 3 A=3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix} \\ & \begin{align*} & \ldots A \mid=3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} -0 \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} +0 \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix} \\ & \quad=3(36-0)=3(36)=108 \end{align*} \end{align*} $$

From equations (i) and (ii), we have:

$|3 A|=27|A|$

Hence, the given result is proved.

Solution

(i) Let $A= \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{vmatrix} $.

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

$|A|=-0 \begin{vmatrix} -1 & -2 \\ -5 & 0\end{vmatrix} +0 \begin{vmatrix} 3 & -2 \\ 3 & 0\end{vmatrix} -(-1) \begin{vmatrix} 3 & -1 \\ 3 & -5\end{vmatrix} =(-15+3)=-12$

(ii) Let $A= \begin{cases} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{cases} $.

By expanding along the first row, we have:

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

(iii) Let $A= \begin{cases} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{cases} $.

By expanding along the first row, we have:

$ \begin{aligned} |A| & =0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} -1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} +2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} \\ & =0-1(0-6)+2(-3-0) \\ & =-1(-6)+2(-3) \\ & =6-6=0 \end{aligned} $

(iv) Let $A= \begin{cases} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{cases} $.

By expanding along the first column, we have:

$ \begin{aligned} |A| & =2 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} -0 \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} +3 \begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix} \\ & =2(0-5)-0+3(1+4) \\ & =-10+15=5 \end{aligned} $

Solution

Let $A= \begin{cases} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{cases} $.

By expanding along the first row, we have:

$ \begin{aligned} |A| & =1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} -1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} -2 \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} \\ & =1(-9+12)-1(-18+15)-2(8-5) \\ & =1(3)-1(-3)-2(3) \\ & =3+3-6 \\ & =6-6 \\ & =0 \end{aligned} $

Solution

(i) $ \begin{vmatrix} 2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix} 2 x & 4 \\ 6 & x\end{vmatrix} $

$\Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4$

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

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

$\Rightarrow x^{2}=3$

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

(ii) $ \begin{vmatrix} 2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix} x & 3 \\ 2 x & 5\end{vmatrix} $

$\Rightarrow 2 \times 5-3 \times 4=x \times 5-3 \times 2 x$

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

$\Rightarrow-2=-x$

$\Rightarrow x=2$

Solution

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

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

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

$\Rightarrow x^{2}=36$

$\Rightarrow x= \pm 6$

Hence, the correct answer is B.