#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 65 sub question (iv) math

Answer$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] and \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]$, such that $A B=A C \ \ but \ \ \ B \neq C, A \neq 0$

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

Solution: Let$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] , \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]$

Here,

$A \neq 0, B \neq C$

Consider LHS

$\\A B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 0 \\ -1 & 0 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$

Now consider RHS

$\\A C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \ \left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0 +0 & 0+0 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$

LHS=RHS

So, $A B=A C \ \ for \ \ \ B \neq C, A \neq 0$

We have $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] and \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]$