Question

$\text{If}A+2B=\left[\begin{array}{cc}2& -4\\ 1& 6\end{array}\right],A^T+B^T=\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right],\text{then}A=$

• A. $\left[\begin{array}{cc}4& 0\\ 3& -8\end{array}\right]$
• B. $\left[\begin{array}{cc}0& 4\\ 3& -8\end{array}\right]$
• C. $\left[\begin{array}{cc}0& 4\\ -8& 3\end{array}\right]$
• D. $\left[\begin{array}{cc}3& 0\\ 3& -8\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}0& 4\\ 3& -8\end{array}\right]$

$A+2B=\left[\begin{array}{cc}2& -4\\ 1& 6\end{array}\right]\cdots \left(i\right)A^T+B^T=\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]{(A^T+B^T)}^T={\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]}^{T}A+B=\left[\begin{array}{cc}1& 0\\ 2& -1\end{array}\right]\cdots \left(ii\right)$
Now, $\left(i\right)-\left(ii\right)$
$B=\left[\begin{array}{cc}2& -4\\ 1& 6\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 2& -1\end{array}\right]B=\left[\begin{array}{cc}1& -4\\ -1& 7\end{array}\right]\therefore A=\left[\begin{array}{cc}1& 0\\ 2& -1\end{array}\right]-\left[\begin{array}{cc}1& -4\\ -1& 7\end{array}\right]A=\left[\begin{array}{cc}0& 4\\ 3& -8\end{array}\right]\therefore A=\left[\begin{array}{cc}0& 4\\ 3& -8\end{array}\right]$