#### Need Solution for R.D.Sharma Maths Class 12 Chapter 4 Algebra of Matrices Exercise Multiple Choice Questions Question 29 Maths Textbook Solution.

Answer: The correct option is $(C) ;-6,-4,-9$

Given:$A=\left[\begin{array}{cc} 0 & 2 \\ 3 & -4 \end{array}\right] \text { and } k A=\left[\begin{array}{cc} 0 & 3 a \\ 2 b & 24 \end{array}\right]$

Solution:

From Matrix,

\begin{aligned} &k A=\left[\begin{array}{cc} 0 & 3 a \\ 2 b & 24 \end{array}\right] \\ &\Rightarrow k A=\left[\begin{array}{cc} 0 & 2 k \\ 3 k & -4 k \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 0 & 2 k \\ 3 k & -4 k \end{array}\right]=\left[\begin{array}{cc} 0 & 3 a \\ 2 b & 24 \end{array}\right] \end{aligned}

Now comparing the Equations,

\begin{aligned} &-4 k=24 \Rightarrow k=-6 \\ &3 k=2 b \Rightarrow 3(-6)=2 b \\ &\Rightarrow 2 b=-18 \\ &\Rightarrow b=-9 \\ &3 a=2 k \Rightarrow 3 a=2(-6) \\ &\Rightarrow 3 a=-12 \\ &\Rightarrow a=-4 \end{aligned}

So the values are k = -6, a= -4 and b = -9

Thus, the correct option is (C).