If $\Delta = \left| {\begin{array}{llllllllllllllllllll}a&p&x\\b&q&y\\c&r&z\end{array}} \right| = 16$, then ${\Delta _1} = \left| {\begin{array}{llllllllllllllllllll}{p + x}&{a + x}&{a + p}\\{q + y}&{b + y}&{b + q}\\{r + z}&{c + z}&{c + r}\end{array}} \right| = 32$.

Correct Answer True

We have,

$\Delta = \left| {\begin{array}{llllllllllllllllllll}a&p&x\\b&q&y\\c&r&z\end{array}} \right| = 16$

and we have to prove,

${\Delta _1} = \left| {\begin{array}{llllllllllllllllllll}{p + x}&{a + x}&{a + p}\\{q + y}&{b + y}&{b + q}\\{r + z}&{c + z}&{c + r}\end{array}} \right| = 32$

${\Delta _1} = \left| {\begin{array}{cccccccccccccccccccc}{2p + 2x + 2a}&{a + x}&{a + p}\\{2q + 2y + 2b}&{b + y}&{b + q}\\{2r + 2z + 2c}&{c + z}&{c + r}\end{array}} \right|$

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

[taking 2 common from ${C_1}$

and then ${C_1} \to {C_1} - {C_2},{C_2} \to {C_2} - {C_3}$]

$= 2\left[ {\begin{array}{llllllllllllllllllll}p&x&{a + p}\\q&y&{b + q}\\r&z&{c + r}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}p&p&{a + p}\\q&q&{b + q}\\r&r&{c + r}\end{array}} \right]$

$= 2\left| {\begin{array}{llllllllllllllllllll}p&x&{a + p}\\q&y&{b + q}\\r&z&{c + r}\end{array}} \right| - 0$

[since, two columns ${C_1}$ and ${C_2}$ are identicals]

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

$= 2\left| {\begin{array}{llllllllllllllllllll}a&p&x\\b&q&y\\c&r&z\end{array}} \right| + 0$

[since, ${C_1}$ and ${C_3}$ are identical in second determinant and in first determinant,

${C_1} \leftrightarrow {C_2}$ and then ${C_1} \leftrightarrow {C_3}$]
$= 2 \times 16$

$= 32$