If A = [3 5], B = [7 3], then find a non-zero matrix

Solution:

Given, A = [3 5]1×2 and B = [7 3]1×2

For AC = BC

We have order of C = 2 x n

For $n=1$, let $C=\left[\begin{array}{l}x \\ y\end{array}\right]$

$\therefore A C=\left[\begin{array}{ll}3 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[3 x+5 y]$

And $B C=\left[\begin{array}{ll}7 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[7 x+3 y]$

For $A C=B C$,

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

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

$4 x=2 y$

$x=\frac{1}{2} y$

$y=2 x$

Hence, $C=\left[\begin{array}{c}x \\ 2 x\end{array}\right]$

It's seen that on taking C of order $2 \times 1,2 \times 2,2 \times 3, \ldots$, we get

$C=\left[\begin{array}{c}x \\ 2 x\end{array}\right],\left[\begin{array}{cc}x & x \\ 2 x & 2 x\end{array}\right],\left[\begin{array}{ccc}x & x & x \\ 2 x & 2 x & 2 x\end{array}\right] \ldots$

In general,

$C=\left[\begin{array}{c}k \\ 2 k\end{array}\right],\left[\begin{array}{cc}k & k \\ 2 k & 2 k\end{array}\right]$ etc $\ldots$