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. -1
• B. 5
• C. 4
• D. 13

Answer 1

The correct option is B 5

Given, $A=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]$ and A adj $A=A{A}^T$
Clearly, A (adj A) = |A| $I_2$
[$\because$ if A is square matrix of order n, then A(adj A) = (adj A) . A = |A| $I_n$]
$=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]I_2=(10a+3b)I_2=(10a+3b)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\left[\begin{array}{cc}10a+3b& 0\\ 0& 10+3b\end{array}\right]$ . . . (i)
and $A{A}^T=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}5a& 3\\ -b& 2\end{array}\right]$
$A{A}^T=\left[\begin{array}{cc}25a^2+b^2& 15a-2b\\ 15a-2b& 13\end{array}\right]$ . . . (ii)
$\because A\left(adjA\right)=A{A}^T$
$\therefore \left[\begin{array}{cc}10a+3b& 0\\ 0& 10a+3b\end{array}\right]=\left[\begin{array}{cc}25a^2+b^2& 15a-2b\\ 15a-2b& 13\end{array}\right]$ using Eqs. (i) and (ii)]
$\Rightarrow$ 15a - 2b = 0
$\Rightarrow a=\frac{2b}{15}$ . . . (iii)
and 10a + 3b = 13 . . . (iv)
On substituting the value of 'a' from Eq. (iii) in Eq. (iv),. we get
$10.\left(\frac{2b}{15}\right)+3b=13\Rightarrow \frac{20b+45b}{15}=13\Rightarrow \frac{65b}{15}=13\Rightarrow b=3$
Now, substituting the value of b in Eq. (iii), we get 5a = 2
Hence, 5a + b = 2 + 3 = 5