If , then find values of x.
$\left[\begin{array}{ccc}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right]$
Given, $\quad\left|\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0$
[Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ], we have
$\Rightarrow \quad\left|\begin{array}{ccc}12+x & 12+x & 12+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0$
Now,
$\left[\right.$ Taking $(12+x)$ common from $\left.R_{1}\right]$
$\Rightarrow \quad(12+x)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & -2 x & 4+x \\ 2 x & 2 x & 4-x\end{array}\right|=0$
$\Rightarrow \quad(12+x)(0-(-2 x)(2 x)]=0$
$(12+x)\left(4 x^{2}\right)=0$
Hence, $\quad x=-12,0$
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