Question

The value of x, for which the matrix A = $\left[\begin{array}{ccc}\frac{2}{x}& -1& 2\\ 1& x& 2x^2\\ 1& \frac{1}{x}& 2\end{array}\right]$ is singular, is

• A. ±1
• B. ±2
• C. ±3
• D. ±4

Answer 1

Answer: (A)

We know that, A is singular if |A| = 0
|A|= A = $\left|\begin{array}{ccc}\frac{2}{x}& -1& 2\\ 1& x& 2x^2\\ 1& \frac{1}{x}& 2\end{array}\right|$ = 0
$\Rightarrow |A|=\frac{2}{x}[2x-2x]+1[2-2x^2]+2[\frac{1}{x}-x]=0$
$\Rightarrow \frac{2}{x}[0]+[2-2x^2]+\frac{2}{x}-2x=0$
$\Rightarrow 2x-2x^3+2-2x^2=0$
$\Rightarrow x^3+x^2-x-1=0$
$\Rightarrow x^2(x+1)-1(x+1))=0$
$\Rightarrow (x+1)(x^2-1)=0$
$\Rightarrow$ x = ±1