Question

If $A$ be a matrix such that $A\times \left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]$ then $A$ is

• A. $\left[\begin{array}{cc}2& 4\\ 1& -1\end{array}\right]$
• B. $\left[\begin{array}{cc}-1& 1\\ 4& 2\end{array}\right]$
• C. $\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]$
• D. none of these

Answer 1

The correct option is C $\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]$

$A\times \left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]$
We will check using options
Option A
Consider, $A\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
$=\left[\begin{array}{cc}2& 4\\ 1& -1\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
$=\left[\begin{array}{cc}6& 12\\ 0& -6\end{array}\right]$
$\ne RHS$
Hence, option $A$ cannot be matrix $A$.
Option B,
Consider, $A\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
$\left[\begin{array}{cc}-1& 1\\ 4& 2\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
$=\left[\begin{array}{cc}0& 6\\ 6& 0\end{array}\right]$
$\ne RHS$
Hence, option $B$ cannot be matrix $A$.
Option C,
Consider, $A\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
$=\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
$=\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]$
$=RHS$
Hence, option C is correct for $A$.