Question

If $\left|\begin{array}{ccc}x& 3& 6\\ 3& 6& x\\ 6& x& 3\end{array}\right|=\left|\begin{array}{ccc}2& x& 7\\ x& 7& 2\\ 7& 2& x\end{array}\right|=\left|\begin{array}{ccc}4& 5& x\\ 5& x& 4\\ x& 4& 5\end{array}\right|=0$, then $x$ is equal to

• A. 0
• B. -9
• C. 3
• D. none of these

Answer 1

The correct option is B -9

$\left|\begin{array}{ccc}x& 3& 6\\ 3& 6& x\\ 6& x& 3\end{array}\right|=\left|\begin{array}{ccc}2& x& 7\\ x& 7& 2\\ 7& 2& x\end{array}\right|=\left|\begin{array}{ccc}4& 5& x\\ 5& x& 4\\ x& 4& 5\end{array}\right|=0$
$C_1\to C_1+C_2+C_3$
$\left|\begin{array}{ccc}x+9& 3& 6\\ x+9& 6& x\\ x+9& x& 3\end{array}\right|=\left|\begin{array}{ccc}x+9& x& 7\\ x+9& 7& 2\\ x+9& 2& x\end{array}\right|=\left|\begin{array}{ccc}x+9& 5& x\\ x+9& x& 4\\ x+9& 4& 5\end{array}\right|=0$
$(x+9)\left|\begin{array}{ccc}1& 3& 6\\ 1& 6& x\\ 1& x& 3\end{array}\right|=(x+9)\left|\begin{array}{ccc}1& x& 7\\ 1& 7& 2\\ 1& 2& x\end{array}\right|=(x+9)\left|\begin{array}{ccc}1& 5& x\\ 1& x& 4\\ 1& 4& 5\end{array}\right|=0$
$R_1\to R_1-R_2,{R_2\to R}_2-R_3$
$(x+9)\left|\begin{array}{ccc}0& -3& 6-x\\ 0& 6-x& x-3\\ 1& x& 3\end{array}\right|=(x+9)\left|\begin{array}{ccc}0& x-7& 5\\ 0& 5& 2-x\\ 1& 2& x\end{array}\right|=(x+9)\left|\begin{array}{ccc}0& 5-x& x-4\\ 0& x-4& -1\\ 1& 4& 5\end{array}\right|=0$
$(x+9)[-3x+9-x^2-36+12x]=(x+9)[-x^2-14+9x-25]=(x+9)[x-5-x^2-16+8x]=0$
$(x+9)[-x^2+9x-27]=(x+9)[-x^2+9x-39]=(x+9)[-x^2-21+9x]=0$
For all quadratics in above equation, $D<0$ , so roots are imaginary
Hence, $x=-9$