Determinants — Class 12 Maths Solution

L.H.S. $=$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{b + c}&{q + r}&{y + z}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|$

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

we get
L.H.S$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2(a + b + c)}&{2(p + q + r)}&{2(x + y + z)}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|$

$= 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{a + b + c}&{p + q + r}&{x + y + z}\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|$

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

we get
L.H.S. $=$ $2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}b&q&y\\{c + a}&{r + p}&{z + x}\\{a + b}&{p + q}&{x + y}\end{array}} \right|$

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

we get
L.H.S. $=$ $2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}b&q&y\\{c + a}&{r + p}&{z + x}\\a&p&x\end{array}} \right|$
Applying${R_2} \to {R_2} - {R_3},$

we get
L.H.S. $=$ $2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}b&q&y\\c&r&z\\a&p&x\end{array}} \right|$

Interchanging ${R_1} \leftrightarrow {R_2},$

we get
L.H.S. $=$ $- 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&p&x\\c&r&z\\b&q&y\end{array}} \right|$

Interchanging ${R_2} \leftrightarrow {R_3},$

we get
LH.S. $=$ $2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&p&x\\b&q&y\\c&r&z\end{array}} \right|$ $=$

R.H.S.