Question

Prove that the roots of the equation $\left|\begin{array}{ccc}a+x& a-x& a-x\\ a-x& a+x& a-x\\ a-x& a-x& a+x\end{array}\right|=0$ are $x=0$ and $x=3a$.

Answer 1

$A=\left|\begin{array}{ccc}a+x& a-x& a-x\\ a-x& a+x& a-x\\ a-x& a-x& a+x\end{array}\right|$

$R_1\to R_1+R_2+R_3$

$A=\left|\begin{array}{ccc}3a-x& 3a-x& 3a-x\\ a-x& a+x& a-x\\ a-x& a-x& a+x\end{array}\right|$

$A=3a-x\left|\begin{array}{ccc}1& 1& 1\\ a-x& a+x& a-x\\ a-x& a-x& a+x\end{array}\right|$

$C_2\to C_1-C_2$

$C_3\to C_1-C_3$

$A=3a-x\left|\begin{array}{ccc}1& 0& 0\\ a-x& -2x& 0\\ a-x& 0& -2x\end{array}\right|$

$\left|A\right|=(3a-x)\times 4x^2=0$

Thus, $x=3a$ and $x=0$ only solutions