Question

On using elementary row operation $R_1 \rightarrow R_1 -3R_2$ in the following matrix equation: $\begin{bmatrix}4&2\\ 3&3\end{bmatrix}=\begin{bmatrix}1&2\\ 0&3\end{bmatrix}\begin{bmatrix}2&0\\ 1&1\end{bmatrix},$ we have

• A. $\begin{bmatrix}-5&-7\\ 3&3\end{bmatrix}=\begin{bmatrix}1&-7\\ 0&3\end{bmatrix}\begin{bmatrix}2&0\\ 1&1\end{bmatrix}$
• B. $\begin{bmatrix}-5&-7\\ 3&3\end{bmatrix}=\begin{bmatrix}1&2\\ 0&3\end{bmatrix}\begin{bmatrix}-1&-3\\ 1&1\end{bmatrix}$
• C. $\begin{bmatrix}-5&-7\\ 3&3\end{bmatrix}=\begin{bmatrix}1&2\\ 1&-7\end{bmatrix}\begin{bmatrix}2&0\\ 1&1\end{bmatrix}$
• D. $\begin{bmatrix}4&2\\ -5&-7\end{bmatrix}=\begin{bmatrix}1&2\\ -3&-3\end{bmatrix}\begin{bmatrix}2&0\\ 1&1\end{bmatrix}$

Answer 1

Answer: (A)

$\begin{bmatrix}4&2\\ 3&3\end{bmatrix}=\begin{bmatrix}1&2\\ 0&3\end{bmatrix}\begin{bmatrix}2&0\\ 1&1\end{bmatrix}$

Applying $R_{1}\rightarrow R_{1}- 3R_{2}$, we get

$\begin{bmatrix}-5&-7\\ 3&3\end{bmatrix}=\begin{bmatrix}1&-7\\ 0&3\end{bmatrix}\begin{bmatrix}2&0\\ 1&1\end{bmatrix}$