Question

What must be the matrix $X$ if $2X+\left[\begin{array}{c}1\\ 3\end{array}\begin{array}{c}2\\ 4\end{array}\right]=\left[\begin{array}{c}3\\ 7\end{array}\begin{array}{c}8\\ 2\end{array}\right]?$

• A. $\left[\begin{array}{c}1\\ 2\end{array}\begin{array}{c}3\\ -1\end{array}\right]$
• B. $\left[\begin{array}{c}1\\ 2\end{array}\begin{array}{c}-3\\ -1\end{array}\right]$
• C. $\left[\begin{array}{c}2\\ 4\end{array}\begin{array}{c}6\\ -2\end{array}\right]$
• D. $\left[\begin{array}{c}2\\ 4\end{array}\begin{array}{c}-6\\ -2\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{c}1\\ 2\end{array}\begin{array}{c}3\\ -1\end{array}\right]$

Since, $2X+\left[\begin{array}{c}1\\ 3\end{array}\begin{array}{c}2\\ 4\end{array}\right]=\left[\begin{array}{c}3\\ 7\end{array}\begin{array}{c}8\\ 2\end{array}\right]$

$\Rightarrow 2X=\left[\begin{array}{cc}3& 8\\ 7& 2\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}2& 6\\ 4& -2\end{array}\right]$

$\Rightarrow X=\frac{1}{2}\left[\begin{array}{cc}2& 6\\ 4& -2\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 2& -1\end{array}\right]$