Question

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

• A. $-1$
• B. $5$
• C. $4$
• D. $13$

Answer 1

The correct option is B $5$

$A.A^T=A.adjA$ ..... $\left(i\right)$

$A.adjA=|A|\left(I\right)$ ........ $\left(ii\right)$ $[A^{-1}=\frac{1}{|A|}adj(A\left)\right]$

Now, $det\left(A\right)=10a+3b$

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

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

$det\left(A\right)|I|=\left[\begin{array}{cc}10a+3b& 0\\ 0& 10a+3b\end{array}\right]$

From $\left(i\right)$ and $\left(ii\right)$

$det\left(A\right)|I|=A.A^T$

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

$⟹10a+3b=13=25a^2+b^2$

And $15a-2b=0$

$⟹15a=2b⟹b=\frac{15a}{2}$

Now, $10a+3b=13$

$⟹20a+45a=26$ .... [By substituting the value of $b$]

$⟹a=\frac{26}{65}$

Now, $b=\frac{15a}{2}=\frac{15\times 26}{2\times 65}=3$

Thus, $5a+b=5\left(\frac{26}{65}\right)+3=5$