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

Answers (1)

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

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