Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.1 Question 21 Maths Textbook Solution.

Answer: $A= B$

Given:
$A= \begin{bmatrix} a+4 &3b \\ 8 &-6 \end{bmatrix},B\begin{bmatrix} 2a+2 &b^{2}+2 \\ 8 & b^{2}-10 \end{bmatrix}$
We have to find out the value of $a$ and $b$
Hint: We will use equality of matrices.
Solution:
$A= \begin{bmatrix} a+4 &3b \\ 8 &-6 \end{bmatrix},B\begin{bmatrix} 2a+2 &b^{2}+2 \\ 8 & b^{2}-10 \end{bmatrix}$

$\because A= B$

Corresponding elements of two equal matrix are equal.

$a+4= 2a+2$                                                                                                                                      ….. (i)

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

$b^{2}-10= -6$                                                                                                                                       ….. (iii)
Solving equation (i), We get

$a+4=2a+2\\\Rightarrow 2a-a=4-2\\\Rightarrow a=2$

Solving equation (ii), We get

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

Solving equation (iii), We get

$b^{2}-10= -6\\\Rightarrow b^{2}= -6+10\\\Rightarrow b^{2}= 4\\\Rightarrow b= \pm 2$

Here, we have common value of $b= 2$ from equation (ii) and (iii)
Hence, $a=2,b= 2$