Determinants — Class 12 Maths Solution

$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x + a}&x&x\\x&{x + a}&x\\x&x&{x + a}\end{array}} \right| = 0$

Applying ${C_1} \to {C_1} + {C_2} + {C_3},$

we get
$\Rightarrow$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{3x + a}&x&x\\{3x + a}&{x + a}&x\\{3x + a}&x&{x + a}\end{array}} \right| = 0$

$\Rightarrow$ $(3x + a)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&x\\1&{x + a}&x\\1&x&{x + a}\end{array}} \right| = 0$

Applying ${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1},$

we get
$\Rightarrow$ $(3x + a)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&x\\0&a&0\\0&0&a\end{array}} \right| = 0$

Applying ${C_2} \to {C_2} - {C_3},$

we get
$\Rightarrow$ $(3x + a)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&x\\0&a&0\\0&{ - a}&a\end{array}} \right| = 0$

Expanding along ${R_1},$

we get $(3x + a)({a^2}) = 0$

Since $a \ne 0 \Rightarrow 3x + a = 0 \Rightarrow x = \cfrac{{ - a}}{3}$ .