Question
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $\text{A}(\text{BC})=(\text{AB})\text{C}.$
$(\text{BC})=\begin{bmatrix}4&0\\1&5\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$
$=\begin{bmatrix}8&0\\7&-10\end{bmatrix}$
and $\text{A}(\text{BC})=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}8&0\\7&-10\end{bmatrix}$ $=\begin{bmatrix}8+14&0-20\\-8+21&0-30\end{bmatrix}=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}$Also, $(\text{AB})=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$
$=\begin{bmatrix}6&10\\-1&15\end{bmatrix}$
$(\text{AB})\text{C}=\begin{bmatrix}6&10\\-1&15\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$
$=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}=\text{A}(\text{BC})$
Hence proved.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Function
$\text{y}=\log\text{x}$