Question

The values of x for which the given matrix $\left[\begin{array}{c}-x\\ 2\\ x\end{array}\begin{array}{c}x\\ x\\ -2\end{array}\begin{array}{c}2\\ -x\\ -x\end{array}\right]$ will be non-singular, are

• A. $-2\le x\le 2$
• B. For all x other than $2$ and $-2$
• C. $x\ge 2$
• D. $x\le -2$

Answer 1

Answer: (B)

For all x other than $2$ and $-2$

Let $\left[\begin{array}{c}-x\\ 2\\ x\end{array}\begin{array}{c}x\\ x\\ -2\end{array}\begin{array}{c}2\\ -x\\ -x\end{array}\right]$ is singular.
$\therefore \left|\begin{array}{c}-x\\ 2\\ x\end{array}\begin{array}{c}x\\ x\\ -2\end{array}\begin{array}{c}2\\ -x\\ -x\end{array}\right|=0$
Applying $C_2\to C_2+C_1,C_3\to C_3+C_1$
$\Rightarrow \left|\begin{array}{c}-x\\ 2\\ x\end{array}\begin{array}{c}0\\ 2+x\\ x-2\end{array}\begin{array}{c}2-x\\ 2-x\\ 0\end{array}\right|=0$
$\Rightarrow (2-x)\left|\begin{array}{c}-x\\ 2\\ x\end{array}\begin{array}{c}0\\ 2+x\\ x-2\end{array}\begin{array}{c}1\\ 1\\ 0\end{array}\right|=0$
Applying $R_2\to R_2-R_1$
$\Rightarrow (2-x)\left|\begin{array}{c}-x\\ 2+x\\ x\end{array}\begin{array}{c}0\\ 2+x\\ x-2\end{array}\begin{array}{c}1\\ 0\\ 0\end{array}\right|=0$
$\Rightarrow (2-x)(2+x)\left|\begin{array}{c}-x\\ 1\\ x\end{array}\begin{array}{c}0\\ 1\\ x-2\end{array}\begin{array}{c}1\\ 0\\ 0\end{array}\right|=0$
$\Rightarrow (2-x)(2+x)(x-2-x)=0$
$\Rightarrow -2(2-x)(2+x)=0$
$\Rightarrow x=2,-2$
$\therefore$ Given matrix is non-singular for all x other than 2 and -2.