Question

If $2A-\left[\begin{array}{cc}1& 2\\ 7& 4\end{array}\right]=\left[\begin{array}{cc}3& 0\\ 0& -2\end{array}\right],$ then $A$ is equal to

• A. $\left[\begin{array}{cc}2& 1\\ 7/2& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}4& 4\\ 7/2& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}3& -1\\ 7& 2\end{array}\right]$
• D. None of these

Answer 1

The correct option is A $\left[\begin{array}{cc}2& 1\\ 7/2& 1\end{array}\right]$

Given, $2A-\left[\begin{array}{cc}1& 2\\ 7& 4\end{array}\right]=\left[\begin{array}{cc}3& 0\\ 0& -2\end{array}\right]$
$\Rightarrow 2A=\left[\begin{array}{cc}3& 0\\ 0& -2\end{array}\right]+\left[\begin{array}{cc}1& 2\\ 7& 4\end{array}\right]=\left[\begin{array}{cc}3+1& 0+2\\ 0+7& -2+4\end{array}\right]=\left[\begin{array}{cc}4& 2\\ 7& 2\end{array}\right]$
Therefore, $A=\left[\begin{array}{cc}2& 1\\ 7/2& 1\end{array}\right]$
Ans: A