Question

If $a^2+b^2+c^2=-2$, and
$A=\left[\begin{array}{ccc}1+a^{2}x& (1+b^2)x& (1+c^2)x\\ (1+a^2)x& 1+b^{2}x& (1+c^2)x\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right]$, then $A$ is non-singular if $x=$

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $0$
B $-1$
D $2$

A is non-singular if $|A|\ne 0$
Consider, $|A|=\left|\begin{array}{ccc}1+a^{2}x& (1+b^2)x& (1+c^2)x\\ (1+a^2)x& 1+b^{2}x& (1+c^2)x\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$

$=\left|\begin{array}{ccc}1+a^{2}x& x+b^{2}x& x+c^{2}x\\ x+a^{2}x& 1+b^{2}x& x+c^{2}x\\ x+a^{2}x& x+b^{2}x& 1+c^{2}x\end{array}\right|$
$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}2x+1+(a^2+b^2+c^2)x& x+b^{2}x& x+c^{2}x\\ 2x+1+(a^2+b^2+c^2)x& 1+b^{2}x& x+c^{2}x\\ 2x+1+(a^2+b^2+c^2)x& x+b^{2}x& 1+c^{2}x\end{array}\right|$
$=\left|\begin{array}{ccc}1& x+b^{2}x& x+c^{2}x\\ 1& 1+b^{2}x& x+c^{2}x\\ 1& x+b^{2}x& 1+c^{2}x\end{array}\right|$ ($\because a^2+b^2+c^2=-2$)
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$\Rightarrow |A|=\left|\begin{array}{ccc}0& x-1& 0\\ 0& -(x-1)& x-1\\ 1& x+b^{2}x& 1+c^{2}x\end{array}\right|$
$\Rightarrow |A|=(x-1{)}^2$

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

$\Rightarrow x\ne 1$

Hence, $A$ is non-singular for all values of $x,$ except $1$