#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 6 Maths Textbook Solution.

Hint:  In LHS, solve (A + B) by adding A and B matrix. Then, add C matrix. and In RHS, first solve (B+C) then add A to it.

Given: To prove:$(A + B) + C = A + (B + C)$

$A = \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} , B = \begin{bmatrix} 9& 7&-1 \\ 3 & 5 & 4\\ 2 & 1 &6 \end{bmatrix} , C = \begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix}$
Solution:
$LHS = (A + B) + C$

$= \left ( \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} + \begin{bmatrix} 9& 7&-1 \\ 3 & 5 & 4\\ 2 & 1 &6 \end{bmatrix} \right )+ \begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix}$

$= \left ( \begin{bmatrix} 2+9 & 1+7 & 1-1\\ 3+3 & -1+5& 0+4\\ 0+2 & 2+1 & 4+6 \end{bmatrix} \right )+ \begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix}$

$= \begin{bmatrix} 2+9+2 & 1+7 -4& 1-1+3\\ 3+3+1 & -1+5-1& 0+4+0\\ 0+2+9 & 2+1 +4& 4+6 +5\end{bmatrix}$

$= \begin{bmatrix} 13 & 4& 3\\ 7 & 3& 4\\ 11 & 7& 15\end{bmatrix}$                                                                                 … (1)

$RHS = A + (B + C)$

$= \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} \left ( \begin{bmatrix} 9& 7&-1 \\ 3 & 5 & 4\\ 2 & 1 &6 \end{bmatrix} +\begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix} \right )$

$= \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} \left ( \begin{bmatrix} 9+2& 7-4&-1+3 \\ 3+1 & 5-1 & 4+0\\ 2+9 & 1+4 &6 +5\end{bmatrix} \right )$

$= \begin{bmatrix} 2+9+2&1+ 7-4&1-1+3 \\3+ 3+1 &-1+0+ 5-1 & 0+4+0\\0+ 2+9 & 2+1+4 &4+6 +5\end{bmatrix}$

$= \begin{bmatrix} 13&4&3 \\7 &3 & 4\\11 & 7 &15\end{bmatrix}$                                                                               … (2)
From (1) and (2)

LHS = RHS

Hence,$(A + B) + C = A + (B + C)$ is proved.