Question

If $\left|\begin{array}{ccc}x& 2& 3\\ 2& 3& x\\ 3& x& 2\end{array}\right|=\left|\begin{array}{ccc}1& x& 4\\ x& 4& 1\\ 4& 1& x\end{array}\right|=\left|\begin{array}{ccc}0& 5& x\\ 5& x& 0\\ x& 0& 5\end{array}\right|=0,$ then the value of x equals $\left(xϵR\right):$

• A. 0
• B. 5
• C. -5
• D. none of these

Answer 1

The correct option is C -5

$\left|\begin{array}{ccc}x& 2& 3\\ 2& 3& x\\ 3& x& 2\end{array}\right|=\left|\begin{array}{ccc}1& x& 4\\ x& 4& 1\\ 4& 1& x\end{array}\right|=\left|\begin{array}{ccc}0& 5& x\\ 5& x& 0\\ x& 0& 5\end{array}\right|=0$
Consider, $\left|\begin{array}{ccc}x& 2& 3\\ 2& 3& x\\ 3& x& 2\end{array}\right|=0$
$\Rightarrow x^3-18x+35=0$
$\Rightarrow (x+5)(x^2-5x+7)=0$
Consider $\left|\begin{array}{ccc}1& x& 4\\ x& 4& 1\\ 4& 1& x\end{array}\right|=0$
$\Rightarrow x^3-12x+65=0$
$\Rightarrow (x+5)(x^2-5x+13)=0$
Consider $\left|\begin{array}{ccc}0& 5& x\\ 5& x& 0\\ x& 0& 5\end{array}\right|=0$
$\Rightarrow x^3=-125$
$\Rightarrow x=-5$
Hence, $x=-5$ is the solution