Question

The value of x, for which the matrix A = $\begin{bmatrix} \frac{2}{x} & -1 & 2 \\ 1 & x &2 x^2 \\ 1 & \frac{1}{x} & 2 \\ \end{bmatrix}$ 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 = $\begin{vmatrix} \frac{2}{x} & -1 & 2 \\ 1 & x &2 x^2 \\ 1 & \frac{1}{x} & 2 \\ \end{vmatrix}$ = 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