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#### Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question (70)

Answer: Hence proved AB=AC, through $B \neq C, A \neq 0$

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

Given

$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right], B=\left[\begin{array}{cc}3 & 1 \\ 5 & 2 \\ -2 & 4\end{array}\right] \quad \mathrm{and} \ \ C=\left[\begin{array}{cc}4 & 2 \\ -3 & 5 \\ 5 & 0\end{array}\right]$

Taking LHS side,

\begin{aligned} A B=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ 5 & 2 \\ -2 & 4\end{array}\right] \\ =\left[\begin{array}{cc}3+5-2 & 1+2+4 \\ 9+15-6 & 3+6+12\end{array}\right] \end{aligned}

$A B=\left[\begin{array}{cc}6 & 7 \\ 18 & 21\end{array}\right] ... (i)$

Now taking RHS side,

\begin{aligned} A C &=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right]\left[\begin{array}{cc}4 & 2 \\ -3 & 5 \\ 5 & 0\end{array}\right] \\ &=\left[\begin{array}{cc}4-3+5 & 2+5+0 \\ 12-9+15 & 6+15+0\end{array}\right] \\ A C &=\left[\begin{array}{cc}6 & 7 \\ 18 & 21\end{array}\right] ... (ii)\end{aligned}

From equation i & ii

AB=AC

Hence, proved