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#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 13 Maths Textbook Solution.

$X = \begin{bmatrix} 12 &\frac{4}{3} \\ \\ 4 &\frac{-14}{3} \\ \\ \frac{25}{3} & \frac{28}{3} \end{bmatrix}$

Hint: Try to separate the ‘X’ variable into LHS.

Given: $A =\begin{bmatrix} 2 &-2 \\ 4 &2 \\ -5 & 1 \end{bmatrix} , B =\begin{bmatrix} 8 & 0\\ 4 & -2\\ 3 & 6 \end{bmatrix} , 2A + 3x = 5B$

Here, we have to compute x.

Solution:

$2A + 3 X = 5B$

$2\begin{bmatrix} 2 & -2\\ 4& 2\\ -5& 1 \end{bmatrix}+3X=5\begin{bmatrix} 8 &0 \\ 4& -2\\ 3 & 6 \end{bmatrix}$

$\begin{bmatrix} 4 & -4\\ 8& 4\\ -10& 2 \end{bmatrix}+3X=\begin{bmatrix} 40 &0 \\ 20& -10\\ 15 & 30 \end{bmatrix}$

$3X=\begin{bmatrix} 40 &0 \\ 20& -10\\ 15 & 30 \end{bmatrix}-\begin{bmatrix} 4 & -4\\ 8& 4\\ -10& 2 \end{bmatrix}$

$3X=\begin{bmatrix} 40-4 &0+4 \\ 20-8& -10-4\\ 15+10 & 30-2 \end{bmatrix}$

$3X=\begin{bmatrix} 36 &4 \\ 12& -14\\ 25 & 28 \end{bmatrix}$

$X=\frac{1}{3}\begin{bmatrix} 36 &4 \\ 12& -14\\ 25 & 28 \end{bmatrix}$

$X = \begin{bmatrix} 12 &\frac{4}{3} \\ \\ 4 &\frac{-14}{3} \\ \\ \frac{25}{3} & \frac{28}{3} \end{bmatrix}$

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