A) 13
B) -1
C) 5
D) 4
Correct Answer: C
Solution :
\[A=\left[ \begin{matrix} 5a & -b \\ 3 & 2 \\ \end{matrix} \right]\]and \[{{A}^{T}}=\left[ \begin{matrix} 5a & 3 \\ -b & 2 \\ \end{matrix} \right]\] \[A{{A}^{T}}=\left[ \begin{matrix} 25{{a}^{2}}+{{b}^{2}} & 15a-2b \\ 15a-2b & 13 \\ \end{matrix} \right]\] Now,\[AadjA=|A|{{I}_{2}}=\left[ \begin{matrix} 10a+3b & 0 \\ 0 & 10a+3b \\ \end{matrix} \right]\] Given \[A{{A}^{T}}=A.\,adj\,A\] 15a ?2b = 0 ...(1) 10a + 3b = 13 ...(2) Solving we get 5a = 2 and b = 3 \[\therefore \]5a + b = 5You need to login to perform this action.
You will be redirected in
3 sec