#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.1 Question 20 Maths Textbook Solution.

Answer: $x= 3,y= 1$

Given:  $\begin{bmatrix} x+10 & y^{2}+2y\\ 0& -4 \end{bmatrix}$$= \begin{bmatrix} 3x+4 & 3\\ 0& y^{2}-5y \end{bmatrix}$

We have to find out the value of $x$  and $y$
Hint: We will use equality of matrices.

Solution:   Here $\begin{bmatrix} x+10 & y^{2}+2y\\ 0& -4 \end{bmatrix}$$= \begin{bmatrix} 3x+4 & 3\\ 0& y^{2}-5y \end{bmatrix}$

Since, corresponding entries of equal matrices are equal. So,

$x+10=3x+4$                                                                                                                                ….. (i)

$y^{2}+2y= 3$                                                                                                                                      ….. (ii)

$-4= y^{2}-5y$                                                                                                                                    ….. (iii)

Solving equation (i), We get

$x+10= 3x+4\\\Rightarrow 2x=6\\\Rightarrow x=3$

Solving equation (ii), We get

$y^{2}+2y-3=0\\\Rightarrow y+3y-y-3=0\\\Rightarrow y\left ( y+3 \right )-1\left ( y+3 \right )=0\\\Rightarrow y=1\: or\: -3$

Solving equation (iii), We get

$-4=y^{2}-5y\\\Rightarrow y^{2}-5y+4=0\\\Rightarrow y^{2}-4y-y+4=0\\\Rightarrow y\left ( y-1 \right )-1( \left ( y-4 \right ) =0\\\Rightarrow \left ( y-1 \right )\left ( y-4 \right )=0\\\Rightarrow y=1\: or\: 4$

From equation (ii) and (iii)
We have common value of $y= 1$
So, $x= 3,y= 1$