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3 and 4 .Determinants and Matrices
easy
यदि $A = \left[ {\begin{array}{*{20}{c}}0&2\\3&{ - 4}\end{array}} \right]$ और $kA = \left[ {\begin{array}{*{20}{c}}0&{3a}\\{2b}&{24}\end{array}} \right]$, तो $k, a, b$ के मान क्रमश: होंगे
A
$ - \,6, - \,12, - \,18$
B
$-6, 4, 9$
C
$ - \,6, - \,4, - \,9$
D
$-6, 12, 18$
Solution
(c)दिया है $kA = \left[ {\,\begin{array}{*{20}{c}}0&{3a}\\{2b}&{24}\end{array}\,} \right]$
==> $k\,\,\left[ {\begin{array}{*{20}{c}}0&2\\3&{ – 4}\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}0&{3a}\\{2b}&{24}\end{array}} \right]$
=> $2k = 3a,\,3k = 2b,\, – 4k = 24$
==> $a = \frac{{2k}}{3},\,\,\,b = \frac{{3k}}{2},\,k = – 6$
$ \Rightarrow $ $k = – 6,\,\,a = – 4,\,b = – 9$.
Standard 12
Mathematics