Question

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

• A. 0
• B. 5
• C. -5
• D. 8

Answer 1

The correct option is C -5

Solving $\left|\begin{array}{ccc}x& 2& 3\\ 2& 3& x\\ 3& x& 2\end{array}\right|=\left|\begin{array}{ccc}0& 5& x\\ 5& x& 0\\ x& 0& 5\end{array}\right|$

$\Rightarrow x(6-x^2)-2(4-3x)+3(2x-9)=0-5(25-0)+x(0-x^2)$

$\Rightarrow 6x-x^3-8+6x+6x-27=-125-x^3$

$\Rightarrow 18x=-90\Rightarrow x=-5$

And solving $\left|\begin{array}{ccc}1& x& 4\\ x& 1& 4\\ 4& 1& x\end{array}\right|=\left|\begin{array}{ccc}0& 5& x\\ 5& x& 0\\ x& 0& 5\end{array}\right|$

$\Rightarrow 1(x-4)-x(x^2-16)+4(x-4)=0-5(25-0)+x(0-x^2)$

$\Rightarrow x-4-x^3+16x+4x-16=125-x^3$

$\Rightarrow 21x=-105\Rightarrow x=-5$

Hence solution is $x=-5$