Question

Let $A=\left[\begin{array}{ccc}1& 1& 1\\ 3& 3& 3\end{array}\right],B=\left[\begin{array}{cc}3& 1\\ 5& 2\\ -2& 4\end{array}\right]\mathrm{and}C=\left[\begin{array}{cc}4& 2\\ -3& 5\\ 5& 0\end{array}\right].$
Verify that AB = AC though B ≠ C, A ≠ O.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& 1& 1\\ 3& 3& 3\end{array}\right],B=\left[\begin{array}{cc}3& 1\\ 5& 2\\ -2& 4\end{array}\right]\mathrm{and}C=\left[\begin{array}{cc}4& 2\\ -3& 5\\ 5& 0\end{array}\right] \\ \mathrm{Now}, \\ AB=\left[\begin{array}{ccc}1& 1& 1\\ 3& 3& 3\end{array}\right]\left[\begin{array}{cc}3& 1\\ 5& 2\\ -2& 4\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}3+5-2& 1+2+4\\ 9+15-6& 3+6+12\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}6& 7\\ 18& 21\end{array}\right] \\ AC=\left[\begin{array}{ccc}1& 1& 1\\ 3& 3& 3\end{array}\right]\left[\begin{array}{cc}4& 2\\ -3& 5\\ 5& 0\end{array}\right] \\ \Rightarrow AC=\left[\begin{array}{cc}4-3+5& 2+5+0\\ 12-9+15& 6+15+0\end{array}\right] \\ \Rightarrow AC=\left[\begin{array}{cc}6& 7\\ 18& 21\end{array}\right] \end{gathered}$$

So, AB = AC though B ≠ C , A ≠ O.