Gujarati
Hindi
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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.