Determinants — Class 12 Maths Solution

We have, $\left| {\begin{array}{cccccccccccccccccccc}x&{x + y}&{x + 2y}\\{x + 2y}&x&{x + y}\\{x + y}&{x + 2y}&x\end{array}} \right|$

$= \left| {\begin{array}{cccccccccccccccccccc}{3(x + y)}&{x + y}&y\\{3(x + y)}&x&y\\{3(x + y)}&{x + 2y}&{ - 2y}\end{array}} \right|$

[and ${C_3} \to {C_3} - {C_2}$]
$= 3(x + y)\left| {\begin{array}{cccccccccccccccccccc}1&{(x + y)}&y\\1&x&y\\1&{(x + 2y)}&{ - 2y}\end{array}} \right|$ [taking $3(x + y)$

common from first column]
$= 3(x + y)\left| {\begin{array}{cccccccccccccccccccc}0&y&0\\1&x&y\\1&{(x + 2y)}&{ - 2y}\end{array}} \right|$

Expanding along ${R_1}$,
$= 3(x + y)[ - y( - 2y - y)]$
$= 3{y^2} \cdot 3(x + y) = 9{y^2}(x + y)$