Determinants — Class 12 Maths Solution

Given, $\left| {\begin{array}{llllllllllllllllllll}{4 - x}&{4 + x}&{4 + x}\\{4 + x}&{4 - x}&{4 + x}\\{4 + x}&{4 + x}&{4 - x}\end{array}} \right| = 0$

$\Rightarrow$ $\left| {\begin{array}{cccccccccccccccccccc}{12 + x}&{12 + x}&{12 + x}\\{4 + x}&{4 - x}&{4 + x}\\{4 + x}&{4 + x}&{4 - x}\end{array}} \right| = 0$

$\Rightarrow$ $(12 + x)\left| {\begin{array}{cccccccccccccccccccc}1&1&1\\{4 + x}&{4 - x}&{4 + x}\\{4 + x}&{4 + x}&{4 - x}\end{array}} \right| = 0$

[taking $(12 + x)$ common from ${R_1}$]

$\Rightarrow$ $(12 + x)\left| {\begin{array}{cccccccccccccccccccc}0&0&1\\0&8&{4 + x}\\{2x}&8&{4 - x}\end{array}} \right| = 0$

and $\left. {{C_2} \to {C_2} + {C_3}} \right]$
$\Rightarrow$ $(12 + x)[1 \cdot ( - 16x)] = 0$

$\Rightarrow$ $(12 + x)( - 16x) = 0$

$\therefore$ $x = - 12,0$