Question
If $\text{A}=\begin{bmatrix}3&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}7&3\end{bmatrix},$ then find a non-zero matrix C such that AC = BC.

Answer

We have, $\text{A}=\begin{bmatrix}3&5\end{bmatrix}_{2\times1}$ and $\text{B}=\begin{bmatrix}7&3\end{bmatrix}_{1\times2}$

Let $\text{C}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}_{2\times1}$ is a non-zero matrix of order 2 × 1.

$\therefore\ \text{AC}=\begin{bmatrix}3&5\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}3\text{x}+5\text{y}\end{bmatrix}$

And $\text{BC}=\begin{bmatrix}7&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}7\text{x}+3\text{y}\end{bmatrix}$

For AC = BC,

$\begin{bmatrix}3\text{x}+5\text{y}\end{bmatrix}=\begin{bmatrix}7\text{x}+3\text{y}\end{bmatrix}$

On using equality of matrix, we get

$\Rightarrow\ 3\text{x}+5\text{y}=7\text{x}+3\text{y}$

$\Rightarrow\ 4\text{x}=2\text{y}$

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

$\Rightarrow\ \text{y}=2\text{x}$

$\therefore\ \text{C}=\begin{bmatrix}\text{x}\\2\text{x}\end{bmatrix}$

We see that on taking C of order 2 × 1, 2 × 2, 2 × 3, .....we get

$\text{C}=\begin{bmatrix}\text{x}\\2\text{x}\end{bmatrix},\begin{bmatrix}\text{x}&\text{x}\\2\text{x}&2\text{x}\end{bmatrix},\begin{bmatrix}\text{x}&\text{x}&\text{x}\\2\text{x}&2\text{x}&2\text{x}\end{bmatrix}....$

In general,

$\text{C}=\begin{bmatrix}\text{k}\\2\text{k}\end{bmatrix},\begin{bmatrix}\text{k}&\text{k}\\2\text{k}&2\text{k}\end{bmatrix}\ \text{etc }.....$

Where, k is any real number.

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