# If $\begin{vmatrix}4+x& 4-x & 4-x\\4-x & 4+x & 4-x\\4-x & 4-x & 4+x\end{vmatrix}$,then find the values of x.

Toolbox:
• If A is a square matrix such that each element of a row (or a column) of A is expressed as the sum of two or more terms,then the determinant can also be expressed as the sum of the determinants of two or more matrices.
• If each element of a row (or a column) of a determinant is multiplied by k,then |A|=k|A|.
Let $\Delta=\begin{vmatrix}4+x & 4-x & 4-x\\4-x & 4+x & 4-x\\4-x & 4-x & 4+x\end{vmatrix}$
Apply $C_1\rightarrow C_1+C_2+C_3$
$\Delta=\begin{vmatrix}12-x & 4-x & 4-x\\12-x & 4+x & 4-x\\12-x & 4-x & 4+x\end{vmatrix}$
Let us take (12-x) as the common factor from $C_1$
$\Delta=(12-x)\begin{vmatrix}1 & 4-x & 4-x\\1 & 4+x & 4-x\\1& 4-x & 4+x\end{vmatrix}$
Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=(12-x)\begin{vmatrix}0 & -2x & 0\\0 & 2x & 2x\\1& 4-x & 4+x\end{vmatrix}$
Now expanding along $R_1$ we get,
$\Delta=(12-x)[0-(-2x)(0+2x)]$
$\quad=(12-x)[2x(4x^2)]$
$\quad=(12-x)(8x^2)$
It is given $|\Delta|=0.$
Therefore $(12-x)(8x^2)=0.$
$\Rightarrow (12-x)=0$ or $(8x^2)=0.$
$\Rightarrow x=12$ or x=0.
Hence the values of x are 0,12.