#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 16 sub question (i) maths textbook solution

Answer: Hence proved $(A B) C=A(B C)$

Hint: Associating property of multiplication is $(A B) C=A(B C)$

Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Prove: $(A B) C=A(B C)$

Given:$A=\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right], B=\left[\begin{array}{cc}1 & 0 \\ -1 & 2 \\ 0 & 3\end{array}\right], C=\left[\begin{array}{c}1 \\ -1\end{array}\right]$

Consider, LHS

$(A B) C=\left(\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 2 \\ 0 & 3\end{array}\right]\right)\left[\begin{array}{c}1 \\ -1\end{array}\right]$

\begin{aligned} &=\left[\begin{array}{cc}1 \times 1+2 \times(-1)+0(0) & 1(0)+2(2)+0(3) \\ -1 \times 1+(-1)(0)+1(0) & (-1)(0)+0(2)+1(3)\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right] \\\\ &=\left[\begin{array}{cc}1-2+0 & 0+4+0 \\ -1+0+0 & 0+0+3\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right] \\\\ &=\left[\begin{array}{cc}-1 & 4 \\ -1 & 3\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right] \\\\(A B) C &=\left[\begin{array}{c}(-1)(1)+4(-1) \\ (-1)(1)+3(-1)\end{array}\right]=\left[\begin{array}{l}-1-4 \\ -1-3\end{array}\right] \\(A B) C &=\left[\begin{array}{l}-5 \\ -4\end{array}\right] & \ldots(i) \end{aligned}

Now consider, RHS

$\begin{array}{l} A(B C)=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left(\left[\begin{array}{cc} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{c} 1 \\ -1 \end{array}\right]\right)\\\\ =\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{c} (1)(1)+0(-1) \\ (-1)(1)+2(-1) \\ (0)(1)+3(-1) \end{array}\right]\\\\ =\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 1+0 \\ -1-2 \\ 0-3 \end{array}\right]\\ \\ =\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ -3 \\ -3 \end{array}\right]\\\\ =\left[\begin{array}{c} (1)(1)+2(-3)+(0)(-3) \\ (-1)(1)+(0)(-3)+(1)(-3) \end{array}\right]=\left[\begin{array}{c} 1-6+0 \\ -1+0-3 \end{array}\right]\\ \\ A(B C)=\left[\begin{array}{l} -5 \\ -4 \end{array}\right] \end{array}$

From equation (i) and (ii), it is clear that  $(A B) C=A(B C)$