Practicing Success
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 : |
$\begin{bmatrix}2 & 3\\4 & 4\end{bmatrix}\begin{bmatrix}-1 & 2\\5 & 0\end{bmatrix}=\begin{bmatrix}2a & b\\2c & d\end{bmatrix}$ $\begin{bmatrix}1 & 3\\2 & 4\end{bmatrix}\begin{bmatrix}-2 & 2\\10 & 0\end{bmatrix}=\begin{bmatrix}2a & b\\2c & d\end{bmatrix}$ $\begin{bmatrix}2 & 3\\4 & 4\end{bmatrix}\begin{bmatrix}-2 & 2\\10 & 0\end{bmatrix}=\begin{bmatrix}2a & b\\2c & d\end{bmatrix}$ $\begin{bmatrix}2 & 3\\4 & 4\end{bmatrix}\begin{bmatrix}-1 & 2\\5 & 0\end{bmatrix}=\begin{bmatrix}2a & 2b\\2c & 2d\end{bmatrix}$ |
$\begin{bmatrix}2 & 3\\4 & 4\end{bmatrix}\begin{bmatrix}-1 & 2\\5 & 0\end{bmatrix}=\begin{bmatrix}2a & b\\2c & d\end{bmatrix}$ |
The correct answer is Option (1) → $\begin{bmatrix}2 & 3\\4 & 4\end{bmatrix}\begin{bmatrix}-1 & 2\\5 & 0\end{bmatrix}=\begin{bmatrix}2a & b\\2c & d\end{bmatrix}$ Using $C_1→2C_1$ $⇒\begin{bmatrix}2×1 & 3\\2×2 & 4\end{bmatrix}\begin{bmatrix}-1 & 2\\5 & 0\end{bmatrix}=\begin{bmatrix}2a & b\\2c & d\end{bmatrix}$ $=\begin{bmatrix}2 & 3\\4 & 4\end{bmatrix}\begin{bmatrix}-1 & 2\\5 & 0\end{bmatrix}=\begin{bmatrix}2a & b\\2c & d\end{bmatrix}$ |