Question

The value of x for which the matrix A = $\begin{bmatrix} 2/x & - 1 & 2 \\ 1 & x & 2x^{2} \\ 1 & 1/x & 2 \end{bmatrix}$is singular is-

• A. ± 1
• B. ± 2
• C. ± 3
• D. None of these

Answer 1

Answer: (A)

± 1

Answer 2

We have

|A| = $\left( \frac{2}{x} \right)$ $\left| \begin{array} { c c } \mathrm { x } & 2 \mathrm { x } ^ { 2 } \\ 1 / \mathrm { x } & 2 \end{array} \right|$ $\left| \begin{matrix} 1 & 2x^{2} \\ 1 & 2 \end{matrix} \right|$+ 2 $\left| \begin{matrix} 1 & x \\ 1 & 1/x \end{matrix} \right|$

= $\frac{2}{x}$ (0) + 2 –2x² + $\left( \frac{1}{x} - x \right)$= $\frac{2x(1 - x^{2}) + 2(1 - x^{2})}{x}$

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

Now, |A| = 0 ̃ x = ± 1