Question

If $A=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]$ and $A$ adj $A=A{A}^T$, then $5a+b$ is equal to:

• A. $13$
• B. $-1$
• C. $5$
• D. $4$

Answer 1

The correct option is C $5$

$A=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]$

$adjA=\left[\begin{array}{cc}2& b\\ -3& 5a\end{array}\right]A^T=\left[\begin{array}{cc}5a& 3\\ -b& 2\end{array}\right]$

According to question,

$AadjA=A{A}^T$

$\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}2& b\\ -3& 5a\end{array}\right]=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}5a& 3\\ -b& 2\end{array}\right]$

$\left[\begin{array}{cc}10a+3b& 0\\ 0& 10a+3b\end{array}\right]=\left[\begin{array}{cc}25a^2+b^2& 15a-2b\\ 15a-2b& 13\end{array}\right]$

On comparing both sides, we get

$15a-2b=0$

$a=\frac{2b}{15}$ ....... $\left(i\right)$

$10a+3b=13$ ...... $\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, we get

$a=\frac{2}{5},b=3$

$5a+b=5\times \frac{2}{5}+3=2+3=5$

$5a+b=5$