#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 18 Subquestion (ii) Maths Textbook Solution.

Answer: $X = \begin{bmatrix} -2 &\frac{-10}{3} \\ \\ 12 &\frac{14}{3} \\ \\ \frac{-31}{3} & \frac{-7}{3} \end{bmatrix}$

Hint: Take the variable 3X into RHS and 5B into LHS. Then, solve.

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

Here, we have to compute X.

Solution:

$2A + 3X = 5B$

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

$\begin{bmatrix} 16 & 0\\ 8& -4\\ 6&12 \end{bmatrix} + 3X = \begin{bmatrix} 10&-10 \\ 20 &10 \\ -25 &5 \end{bmatrix}$

$3X = \begin{bmatrix} 10&-10 \\ 20 &10 \\ -25 &5 \end{bmatrix}-\begin{bmatrix} 16 & 0\\ 8& -4\\ 6&12 \end{bmatrix}$

$3X = \begin{bmatrix} -6 &-10 \\ 12 &14 \\ -31 &-7 \end{bmatrix}$

$X = \frac{1}{3}\begin{bmatrix} -6 &-10 \\ 12 &14 \\ -31 &-7 \end{bmatrix}$

$X = \begin{bmatrix} -2 &\frac{-10}{3} \\ \\ 12 &\frac{14}{3} \\ \\ \frac{-31}{3} & \frac{-7}{3} \end{bmatrix}$