If the elementary column transformation $C_1→2C_1$ is carried out on the following matrix equation.

$\begin{bmatrix}1  & 3\\2 & 4\end{bmatrix}\begin{bmatrix}-1  & 2\\5 & 0\end{bmatrix}=\begin{bmatrix}a  & b\\c & d\end{bmatrix}$ Then the transformed equation will be :

$\begin{bmatrix}1  & 3\\2 & 4\end{bmatrix}\begin{bmatrix}-2  & 2\\10 & 0\end{bmatrix}=\begin{bmatrix}2a  & b\\2c & d\end{bmatrix}$

The correct answer is Option (2) → $\begin{bmatrix}1  & 3\\2 & 4\end{bmatrix}\begin{bmatrix}-2  & 2\\10 & 0\end{bmatrix}=\begin{bmatrix}2a  & b\\2c & d\end{bmatrix}$

Here, the operation is $C_1 \rightarrow 2C_1$, which means the first column is multiplied by 2.

Apply this to the second matrix:

$\begin{bmatrix} -1 & 2 \\ 5 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} -2 & 2 \\ 10 & 0 \end{bmatrix}$

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \rightarrow \begin{bmatrix} 2a & b \\ 2c & d \end{bmatrix}$

The first matrix remains unchanged:

$\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

Therefore, the transformed equation becomes:

$\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 2 \\ 10 & 0 \end{bmatrix} = \begin{bmatrix} 2a & b \\ 2c & d \end{bmatrix}$