Suppose D = left| {,begin{array}{*{20}{c}}{{a₁}}&{{b₁}}&{{c₁}}{{a₂}}&{{b₂}}&am

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3 and 4 .Determinants and Matrices

hard

Suppose $D = \left| {\,\begin{array}{{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ 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

$D' = D(1 - pqr)$

C

$D' = D(1 + p + q + r)$

D

$D' = D(1 + pqr)$

Solution

(d) $D' = D + pqr\,D = D(1 + pqr)$.

Trick : Check by putting $a_1 = b_2 = c_3 = 1$ and all other zero.

Standard 12

Mathematics

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