#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question15 Subquestion (i) Maths Textbook Solution.

Answer: $x = 32\: and\: y = -32$

Hint: Solve LHS and then equate with RHS.

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

Here, we have to compute x  and y.

Solution:

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

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

Equating this, we get:

$x-y + 3 = 6$
$\Rightarrow x-y = 3$                                                                                                       … (1)

Also,

$x = 2x + y$
$\Rightarrow y = -x$                                                                                                …. (2)

Applying the value of ‘y’ in equation (1), we get:

$x-(-x) = 3$
$2x = 3$
$x = 32$

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

$y = -32$

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Answer:  $x = \frac{3}{2} \: and\: y = -\frac{-3}{2}$
Hint: Solve LHS and then equate with RHS.

Given:$\begin{bmatrix} x-y& 2 &-2 \\ 4 & x & 6 \end{bmatrix}+\begin{bmatrix} 3 &-2 &2 \\ 1&0 & -1 \end{bmatrix}= \begin{bmatrix} 6 & 0 &0 \\ 5 & 2x+y & 5 \end{bmatrix}$
Here, we have to compute x  and y.
Solution:

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

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

Equating this, we get:

$x-y + 3 = 6$
$\Rightarrow x-y = 3$                                                                                                               … (1)

Also,

$x = 2x + y$
$\Rightarrow y = -x$                                                                                                 …. (2)

Applying the value of ‘y’ in equation (1), we get:

$x-(-x) = 3$
$2x = 3$
$x = \frac{3}{2}$

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

$y = -\frac{3}{2}$