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3 and 4 .Determinants and Matrices
medium
$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $
A
$1$
B
$0$
C
$3$
D
$a + b + c$
Solution
(b) ${C_3} \to {C_3} – {C_1}$और ${C_2} \to {C_2} – {C_1}$,
के द्वारा $\left| {\,\begin{array}{*{20}{c}}1&{ac}&{bc}\\1&{ad}&{bd}\\1&{ae}&{be}\end{array}\,} \right| = ab\left| {\,\begin{array}{*{20}{c}}1&c&c\\1&d&d\\1&e&e\end{array}\,} \right| = 0$,.
$\{ \because {C_2} \equiv {C_3}\} $
Standard 12
Mathematics