#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 10 Maths Textbook Solution.

Answer: $X = \begin{bmatrix} 2 &3 &1 \\ 0 &1 &2 \\ 6 &4 &0 \end{bmatrix} and Y = \begin{bmatrix} 1 & 2 &0 \\ -1 & 0 &2 \\ 5 & 4 & 0 \end{bmatrix}$

Hint: Try to add two matrices and find variable ‘X ’. Then substitute ‘X ’ value in any matrix to find ‘Y ’.

Given:$X+Y= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix} and X-Y = \begin{bmatrix} 1 &1 &1 \\ 1 &1 &0 \\ 1&0 &0 \end{bmatrix}$

Here, we have to compute X  and Y .

Solution:

$X+Y+X-Y = \begin{bmatrix} 3 & 5 &1 \\ -1 &1 &4 \\ 11 & 8 &0 \end{bmatrix} +\begin{bmatrix} 1 & 1 & 1\\ 1 & 1& 0\\ 1 &0 & 0 \end{bmatrix}$

$X+Y+X-Y= \begin{bmatrix} 3+1 & 5+1& 1+1\\ -1+1 &1+1 &4+0 \\ 11+1 & 8+0 & 0+0 \end{bmatrix}$

$2X= \begin{bmatrix} 4 & 6& 2\\ 0 &2 &4\\ 12 & 8 & 0 \end{bmatrix}$

$X= \frac{1}{2}\begin{bmatrix} 4 & 6& 2\\ 0 &2 &4\\ 12 & 8 & 0 \end{bmatrix}$

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

Now,

$X+Y= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}$

$\begin{bmatrix} 2 & 3& 1\\ 0 &1 &2\\ 6 & 4 & 0 \end{bmatrix}+Y$  $= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}$$= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}$

$Y = \begin{bmatrix} 3-2 & 5-3 & 1-1\\ -1-0 & 1-1 &4-2 \\ 11-6 &8-4 & 0-0 \end{bmatrix}$

$Y = \begin{bmatrix} 1 & 2 &0 \\ -1 & 0 &2 \\ 5 & 4 & 0 \end{bmatrix}$