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

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