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3 and 4 .Determinants and Matrices
normal
If $D = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|$ and $D' = \left| {\begin{array}{*{20}{c}} {{a_1} + p{b_1}}&{{b_1} + q{c_1}}&{{c_1} + r{a_1}} \\ {{a_2} + p{b_2}}&{{b_2} + q{c_2}}&{{c_2} + r{a_2}} \\ {{a_3} + p{b_3}}&{{b_3} + q{c_3}}&{{c_3} + r{a_3}} \end{array}} \right|$ then
A
$D' = D$
B
$D' = (1 + pqr)D$
C
$D' = (1 -pqr)D$
D
$D' = pqrD$
Solution
$D = \left| {\begin{array}{*{20}{c}}
{{a_1} + p{b_1}}&{{b_1} + q{c_1}}&{r{a_1}} \\
{{a_2} + p{b_2}}&{{b_2} + q{c_2}}&{r{a_2}} \\
{{a_3} + p{b_3}}&{{b_3} + q{c_3}}&{r{a_3}}
\end{array}} \right| = D'$
$\Rightarrow$ $D' = (1 + pqr) D$
Standard 12
Mathematics