# Solve this

Question:

If $\left|\begin{array}{lll}2-x & 2+x & 2+x \\ 2+x & 2-x & 2+x \\ 2+x & 2+x & 2-x\end{array}\right|=0$, then $x=$

Solution:

Given: $\left|\begin{array}{lll}2-x & 2+x & 2+x \\ 2+x & 2-x & 2+x \\ 2+x & 2+x & 2-x\end{array}\right|=0$

$\left|\begin{array}{lll}2-x & 2+x & 2+x \\ 2+x & 2-x & 2+x \\ 2+x & 2+x & 2-x\end{array}\right|=0$

Applying $R_{2} \rightarrow R_{2}-R_{1}$

$\Rightarrow\left|\begin{array}{ccc}2-x & 2+x & 2+x \\ 2+x-2+x & 2-x-2-x & 2+x-2-x \\ 2+x & 2+x & 2-x\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}2-x & 2+x & 2+x \\ 2 x & -2 x & 0 \\ 2+x & 2+x & 2-x\end{array}\right|=0$

Taking $(2 x)$ common from $R_{2}$

$\Rightarrow(2 x)\left|\begin{array}{ccc}2-x & 2+x & 2+x \\ 1 & -1 & 0 \\ 2+x & 2+x & 2-x\end{array}\right|=0$

Applying $R_{3} \rightarrow R_{3}-R_{1}$

$\Rightarrow(2 x)\left|\begin{array}{ccc}2-x & 2+x & 2+x \\ 1 & -1 & 0 \\ 2+x-2+x & 2+x-2-x & 2-x-2-x\end{array}\right|=0$

$\Rightarrow(2 x)\left|\begin{array}{ccc}2-x & 2+x & 2+x \\ 1 & -1 & 0 \\ 2 x & 0 & -2 x\end{array}\right|=0$

Taking $(2 x)$ common from $R_{3}$

$\Rightarrow(2 x)^{2}\left|\begin{array}{ccc}2-x & 2+x & 2+x \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{array}\right|=0$

$\Rightarrow(2 x)^{2}\left|\begin{array}{ccc}2-x & 2+x & 4 \\ 1 & -1 & 1 \\ 1 & 0 & 0\end{array}\right|=0$

Expanding through $R_{3}$

$\Rightarrow(2 x)^{2}[(1)((2+x)(1)+4)]=0$

$\Rightarrow(2 x)^{2}[(2+x+4)]=0$

$\Rightarrow(2 x)^{2}(x+6)=0$

$\Rightarrow(2 x)^{2}=0$ or $(6+x)=0$

$\Rightarrow x=0$ or $x=-6$

Hence, $x=\underline{0}, \underline{-6}$.