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3 and 4 .Determinants and Matrices
easy
यदि $a \ne 6,b,c$ सारणिक $\left| {\,\begin{array}{*{20}{c}}a&{2b}&{2c}\\3&b&c\\4&a&b\end{array}\,} \right| = 0,$ तो $abc = $
A
$a + b + c$
B
$0$
C
${b^3}$
D
$ab + bc$
Solution
(c) $\left| {\,\begin{array}{*{20}{c}}
a&{2b}&{2c} \\
3&b&c \\
4&a&b
\end{array}\,} \right| = 0 = > \left| {\,\begin{array}{*{20}{c}}
{a – 6}&0&0 \\
3&b&c \\
4&a&b
\end{array}\,} \right| = 0$ $[R_1 → R_1 -2R_2]$
==> $(a – 6)({b^2} – ac) = 0 \Rightarrow {b^2} – ac = 0$ $a \ne 6$
$\therefore $ $ac = {b^2} \Rightarrow abc = {b^3}.$
Standard 12
Mathematics
