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 :
Answer & explanation
Correct answer: option 2
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:
$\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 2 \\ 10 & 0 \end{bmatrix} = \begin{bmatrix} 2a & b \\ 2c & d \end{bmatrix}$