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

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

Answers (1)

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)

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