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3 and 4 .Determinants and Matrices
normal
$\left| {\,\begin{array}{*{20}{c}}{13}&{16}&{19}\\{14}&{17}&{20}\\{15}&{18}&{21}\end{array}\,} \right| = $
A
$0$
B
$-39$
C
$96$
D
$57$
Solution
(a) ${C_3} \to {C_3} – {C_2}$ और ${C_2} \to {C_2} – {C_1}$
के द्वारा, $\left| {\,\begin{array}{*{20}{c}}{13}&3&3\\{14}&3&3\\{15}&3&3\end{array}\,} \right| = 0$ ,
$\{ \because {C_2} \equiv {C_3}\} $
Standard 12
Mathematics