#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 15 Subquestion (iii) Maths Textbook Solution.

Answer: $x = 1, y = 2$
Hint: Solve LHS part and then equate with RHS part.
Given:

$x \begin{bmatrix} 2\\1 \end{bmatrix} + y \begin{bmatrix} 3\\5 \end{bmatrix} +\begin{bmatrix} -8\\11 \end{bmatrix} = 0$
Here, we have to compute x  and y .
Solution:

$x \begin{bmatrix} 2\\1 \end{bmatrix} + y \begin{bmatrix} 3\\5 \end{bmatrix} +\begin{bmatrix} -8\\11 \end{bmatrix} = 0$

$\begin{bmatrix} 2x+3y-8\\x+5y-11 \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix}$

Equating this, we get:

$2x+3y-8 = 0$
$\Rightarrow 2x+3y=8$                                                                                      …(1)

Also,

$x+5y-11 = 0$
$\Rightarrow x=11-5y$                                                                                      …(2)
Applying the value of ‘x ’ in equation (1), we get:

$2(11 -5y) + 3y = 8$
$22 -10y + 3y = 8$
$-7y = 8 -22$
$-7y = -14$
$y = 2$

Applying the value of ‘y’ in equation (2), we get:

$x = 11 - 5(2)$
$x = 11- 10$
$x = 1$