#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 12 Maths Textbook Solution.

Answer:  $C = \begin{bmatrix} -24 &-10 \\ -28 & -38 \end{bmatrix}$

Hint: Null matrix is $\begin{bmatrix} 0 &0\\ 0 & 0 \end{bmatrix}$

Given:   $A =\begin{bmatrix} 9 &1 \\ 7 & 8 \end{bmatrix} , B =\begin{bmatrix} 1 & 5\\ 7 & 12 \end{bmatrix}, 5A + 3B + 2C = 0$

Here, we have to compute C.

Solution:

$5A + 3B + 2C = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

$5\begin{bmatrix} 9 & 1\\ 7 & 8 \end{bmatrix} + 3\begin{bmatrix} 1 & 5\\ 7 & 12 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$

$\begin{bmatrix} 45 & 5\\ 35 & 40 \end{bmatrix} + \begin{bmatrix} 3 & 15\\ 21 & 36 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$

$\begin{bmatrix} 45+3 & 5+15\\ 35+21 & 40+36 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$

$\begin{bmatrix} 48 & 20\\ 56 & 76 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$

$2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}-\begin{bmatrix} 48 & 20\\ 56 & 76 \end{bmatrix}$

$2C = \begin{bmatrix} -48 & -20\\ -56 & -76 \end{bmatrix}$

$C =\frac{1}{2}\begin{bmatrix} -48 & -20\\ -56 & -76 \end{bmatrix}$

$C = \begin{bmatrix} -24 &-10 \\ -28 & -38 \end{bmatrix}$