If $A = [3\,\,\,\,5]$ and $B = [7\,\,\,\,3]$, then find a non-zero matrix $C$ such that $AC = BC$.

We have, $A = {[3\,\,\,\,5]_{1 \times 2}}$ and $B = {[7\,\,\,\,3]_{1 \times 2}}$

Let $C = {\left[ {\begin{array}{llllllllllllllllllll}x\\y\end{array}} \right]_{2 \times 1}}$

is a non-zero matrix of order $2 \times 1$.
$\therefore$ $AC = \left[ {\begin{array}{llllllllllllllllllll}3&5\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}x\\y\end{array}} \right] = [3x + 5y]$
and $BC$ $ = \left[ {\begin{array}{llllllllllllllllllll}7&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}x\\y\end{array}} \right] = [7x + 3y]$
For $AC = BC$

$[3x + 5y] = [7x + 3y]$

On using equality of matrix,

we get
$3x + 5y = 7x + 3y$
$\Rightarrow$ $4x = 2y$
$\Rightarrow$ $x = \frac{1}{2}y$

$\Rightarrow$ $y = 2x$
$\therefore$ $C = \left[ {\begin{array}{cccccccccccccccccccc}x\\{2x}\end{array}} \right]$

We see that on taking $C$ of order $2 \times 1,2 \times 2,2 \times 3, \ldots$,

we get $C = \left[ {\begin{array}{cccccccccccccccccccc}x\\{2x}\end{array}} \right],\left[ {\begin{array}{cccccccccccccccccccc}x&x\\{2x}&{2x}\end{array}} \right],\left[ {\begin{array}{cccccccccccccccccccc}x&x&x\\{2x}&{2x}&{2x}\end{array}} \right] \ldots$

In general,
$C = \left[ {\begin{array}{cccccccccccccccccccc}k\\{2k}\end{array}} \right],\left[ {\begin{array}{cccccccccccccccccccc}k&k\\{2k}&{2k}\end{array}} \right]$ etc $\ldots$

where, $k$ is any real number.