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Solve the equations $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0, a \neq 0$
$x=-\frac{a}{3}$
$x=-\frac{a}{6}$
$x=-\frac{a}{2}$
$x=-\frac{a}{4}$
Solution
$\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we get:
$\left|\begin{array}{ccc}3 x+a & 3 x+a & 3 x+a \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$
$\Rightarrow(3 x+a)\left|\begin{array}{ccc}1 & 1 & 1 \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:
$\Rightarrow(3 x+a)\left|\begin{array}{lll}1 & 1 & 1 \\ x & a & x \\ x & x & a\end{array}\right|=0$
Expanding along $R_{1},$ we have:
$(3 x+a)\left[1 \mathrm{x} a^{2}\right]=0$
$\Rightarrow a^{2}(3 x+a)=0$
But $a \neq 0$
Therefore, we have:
$3 x+a=0$
$\Rightarrow x=-\frac{a}{3}$
