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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
Question
$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $
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Answer

b
(b) Applying ${C_3} \to {C_3} - {C_1}$and ${C_2} \to {C_2} - {C_1}$,

we get $\left| {\,\begin{array}{*{20}{c}}1&{ac}&{bc}\\1&{ad}&{bd}\\1&{ae}&{be}\end{array}\,} \right| = ab\left| {\,\begin{array}{*{20}{c}}1&c&c\\1&d&d\\1&e&e\end{array}\,} \right| = 0$,. 

$\{ \because {C_2} \equiv {C_3}\} $

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