For what value of $x$ the matrix $A$ is singular?

Solution:

(i) Matrix $A$ will be singular if

$|A|=0$

$|A|=\left|\begin{array}{ll}1+x & 7 \\ 3-x & 8\end{array}\right|=0$

$\Rightarrow 8+8 x-21+7 x=0$

$\Rightarrow 15 x-13=0$

$\Rightarrow 15 x=13$

$\Rightarrow x=\frac{13}{15}$

(ii) Matrix A will be singular if

$|A|=0$

$\Rightarrow(x-1)\left[(x-1)^{2}-1\right]-1(x-1-1)+1[1-(x-1)]=0$

$\Rightarrow(x-1)\left(x^{2}-2 x\right)-1(x-2)+1(2-x)=0$

$\Rightarrow x^{3}-2 x^{2}-x^{2}+2 x-x+2-x+2=0$

$\Rightarrow x^{3}-3 x^{2}+4=0$

$\Rightarrow(x-2)^{2}(x+1)=0$

$\Rightarrow x=2$ or $x=-1$