Question

$If$ $2A+B^T=\left[\begin{array}{cc}2& -3\\ 4& 7\end{array}\right]$ and $A^T-B=\left[\begin{array}{cc}4& -5\\ 0& 1\end{array}\right]$,then $A=$

• A. $\frac{1}{3}\left[\begin{array}{cc}6& 3\\ -1& 8\end{array}\right]$
• B. $\left[\begin{array}{cc}2& 3\\ -1& 8\end{array}\right]$
• C. $\frac{1}{2}\left[\begin{array}{cc}2& 3\\ -1& 8\end{array}\right]$
• D. $0$

Answer 1

Answer: (A)

$\frac{1}{3}\left[\begin{array}{cc}6& 3\\ -1& 8\end{array}\right]$

Given: $2A+B^T=\left[\begin{array}{cc}2& -3\\ 4& 7\end{array}\right]$

and $A^T-B=\left[\begin{array}{cc}4& -5\\ 0& 1\end{array}\right]$

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Then $2A+B^T=\left[\begin{array}{cc}2& -3\\ 4& 7\end{array}\right]$ implies

$\Rightarrow \left[\begin{array}{cc}2a& 2b\\ 2c& 2d\end{array}\right]+B^T=\left[\begin{array}{cc}2& -3\\ 4& 7\end{array}\right]$

$\Rightarrow B^T=\left[\begin{array}{cc}2& -3\\ 4& 7\end{array}\right]-\left[\begin{array}{cc}2a& 2b\\ 2c& 2d\end{array}\right]$

$\Rightarrow B^T=\left[\begin{array}{cc}2-2a& -3-2b\\ 4-2c& 7-2d\end{array}\right]$

$\Rightarrow B=\left[\begin{array}{cc}2-2a& 4-2c\\ -3-2b& 7-2d\end{array}\right]$

Put the matrix $B$ in $A^T-B=\left[\begin{array}{cc}4& -5\\ 0& 1\end{array}\right]$

$\Rightarrow A^T=B+\left[\begin{array}{cc}4& -5\\ 0& 1\end{array}\right]$

$\Rightarrow A^T=\left[\begin{array}{cc}2-2a& 4-2c\\ -3-2b& 7-2d\end{array}\right]+\left[\begin{array}{cc}4& -5\\ 0& 1\end{array}\right]$

$\Rightarrow A^T=\left[\begin{array}{cc}6-2a& -1-2c\\ -3-2b& 8-2d\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}a& c\\ b& d\end{array}\right]=\left[\begin{array}{cc}6-2a& -1-2c\\ -3-2b& 8-2d\end{array}\right]$

$\therefore a=6-2a,1-2c=c,-3-2b=b,8-2d=d$

$\Rightarrow a=3,c=-\frac{1}{3},b=1,d=\frac{8}{3}$

So $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}2& -1\\ -\frac{1}{3}& \frac{8}{3}\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}6& -3\\ -1& 8\end{array}\right]$

Answer: option A is correct