Determinants — Class 12 Maths Solution

Applying ${R_1} \leftrightarrow {R_2},{R_2} \leftrightarrow {R_3},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - a}&0&{ - c}\\b&c&0\\0&a&{ - b}\end{array}} \right|$

Applying ${R_1} \leftrightarrow \cfrac{{ - {R_1}}}{a},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{c/a}\\b&c&0\\0&a&{ - b}\end{array}} \right|$

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

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{c/a}\\0&c&{ - bc/a}\\0&a&{ - b}\end{array}} \right|$

Applying${R_2} \to \cfrac{{{R_2}}}{c},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{c/a}\\0&1&{ - b/a}\\0&a&{ - b}\end{array}} \right|$

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

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{c/a}\\0&1&{ - b/a}\\0&0&0\end{array}} \right| = 0$

(since each element of ${R_3}$ is 0)