Question

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

Answer 1

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



$$\begin{gathered} \left|\begin{array}{ccc}2-x& 2+x& 2+x\\ 2+x& 2-x& 2+x\\ 2+x& 2+x& 2-x\end{array}\right|=0 \\ \mathrm{Applying}{R}_2\to R_2-R_1 \\ \Rightarrow \left|\begin{array}{ccc}2-x& 2+x& 2+x\\ 2+x-2+x& 2-x-2-x& 2+x-2-x\\ 2+x& 2+x& 2-x\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}2-x& 2+x& 2+x\\ 2x& -2x& 0\\ 2+x& 2+x& 2-x\end{array}\right|=0 \\ \mathrm{Taking}\left(2x\right)\mathrm{common}\mathrm{from}{R}_2 \\ \Rightarrow \left(2x\right)\left|\begin{array}{ccc}2-x& 2+x& 2+x\\ 1& -1& 0\\ 2+x& 2+x& 2-x\end{array}\right|=0 \\ \mathrm{Applying}{R}_3\to R_3-R_1 \\ \Rightarrow \left(2x\right)\left|\begin{array}{ccc}2-x& 2+x& 2+x\\ 1& -1& 0\\ 2+x-2+x& 2+x-2-x& 2-x-2-x\end{array}\right|=0 \\ \Rightarrow \left(2x\right)\left|\begin{array}{ccc}2-x& 2+x& 2+x\\ 1& -1& 0\\ 2x& 0& -2x\end{array}\right|=0 \\ \mathrm{Taking}\left(2x\right)\mathrm{common}\mathrm{from}{R}_3 \\ \Rightarrow {\left(2x\right)}^2\left|\begin{array}{ccc}2-x& 2+x& 2+x\\ 1& -1& 0\\ 1& 0& -1\end{array}\right|=0 \\ \mathrm{Applying}{C}_3\to C_3+C_1 \\ \Rightarrow {\left(2x\right)}^2\left|\begin{array}{ccc}2-x& 2+x& 2+x+2-x\\ 1& -1& 0+1\\ 1& 0& -1+1\end{array}\right|=0 \\ \Rightarrow {\left(2x\right)}^2\left|\begin{array}{ccc}2-x& 2+x& 4\\ 1& -1& 1\\ 1& 0& 0\end{array}\right|=0 \\ \mathrm{Expanding}\mathrm{through}{R}_3 \\ \Rightarrow {\left(2x\right)}^2\left[\left(1\right)\left(\left(2+x\right)\left(1\right)+4\right)\right]=0 \\ \Rightarrow {\left(2x\right)}^2\left[\left(2+x+4\right)\right]=0 \\ \Rightarrow {\left(2x\right)}^2\left(x+6\right)=0 \\ \Rightarrow {\left(2x\right)}^2=0\mathrm{or}\left(6+x\right)=0 \\ \Rightarrow x=0\mathrm{or}x=-6 \end{gathered}$$

Hence, x = 0, −6.