MCQ
Choose the correct answer from the given four options. On using elementary column operations $C_2 \rightarrow C_2-2 C_1$ in the following matrix equation $\begin{bmatrix}1&-3\\2&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&1\\2&4\end{bmatrix},$ we have :
  • A
    $\begin{bmatrix}1&-5\\0&4\end{bmatrix}=\begin{bmatrix}1&-1\\-2&2\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
  • B
    $\begin{bmatrix}1&-5\\0&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\-0&2\end{bmatrix}$
  • C
    $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-3\\0&1\end{bmatrix}\begin{bmatrix}3&1\\-2&4\end{bmatrix}$
  • $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$

Answer

Correct option: D.
$\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
Given that, $\begin{bmatrix}1&-3\\2&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&1\\2&4\end{bmatrix}$
On using $C_2 \rightarrow C_2-2 C_1$ $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
Since, on using elementary column operation on $X = AB,$
we apply these operations simultaneously on $X$ and on the second matrix $B$ of the product $AB$ on $\text{RHS}.$

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