Matrices and Determinants 3 Question 5
6. If $A=\begin{bmatrix}5 a & -b \\ 3 & 2\end{bmatrix}$ and $A$ adj $A=A A^{T}$, then $5 a+b$ is equal to
(2016 Main)
(a) -1
(b) 5
(c) 4
(d) 13
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Answer:
Correct Answer: 6. (b)
Solution:
- Given, $A=\begin{bmatrix}5 a & -b \\ 3 & 2\end{bmatrix}$ and $A$ adj $A=A A^{T}$
Clearly, $A(\operatorname{adj} A)=|A| I_{2}$
$ [\because \text { if } A \text { is square matrix of order } n $ then $\left.A(\operatorname{adj} A)=(\operatorname{adj} A) \cdot A=|A| I_{n}\right]$
$ =\begin{bmatrix} 5 a & -b \\ 3 & 2 \end{bmatrix} $ $I_{2}=(10 a+3 b) I_{2} $
$=(10 a+3 b)$ $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 10 a+3 b & 0 \\ 0 & 10 a +3 b \end{bmatrix} $
and $A A^{T}=\begin{bmatrix} 5 a & -b & 5 a & 3 \\ 3 & 2 & -b & 2 \end{bmatrix} $
= $\begin{bmatrix} 25 a^{2}+ b^{2} & 15 a-2 b \\ 15 a-2 b & 13 \end{bmatrix} $
$\because \quad A(\operatorname{adj} A)=A A^{T} $
$\therefore \begin{bmatrix} 10 a+3 b & 0 \\ 0 & 10 a+3 b \end{bmatrix}$ = $\begin{bmatrix} 25 a^{2}+b^{2} & 15 a-2 b \\ 15 a-2 b & 13 \end{bmatrix} $
$\Rightarrow \quad 15 a-2 b=0 $
$ \Rightarrow \quad a=\frac{2 b}{15}$
and $10a + 3b = 13$
On substituting the value of ’ $a$ ’ from Eq. (iii) in Eq. (iv), we get
$ 10 \cdot \frac{2 b}{15}+3 b =13 $
$\Rightarrow \frac{20 b+45 b}{15} =13 $
$\Rightarrow \frac{65 b}{15} =13 $
$\Rightarrow b =3$
Now, substituting the value of $b$ in Eq. (iii), we get
$5 a=2$
Hence, $\quad 5 a+b=2+3=5$
