Suppose A is any $3 \times 3$ non-singular matrix and $(\mathrm{A}-3 \mathrm{I})(\mathrm{A}-5 \mathrm{I})=0$…
Suppose A is any $3 \times 3$ non-singular matrix and $(\mathrm{A}-3 \mathrm{I})(\mathrm{A}-5 \mathrm{I})=0$ where $\mathrm{I}=\mathrm{I}_3$ and $\mathrm{O}=\mathrm{O}_3$. Here $\mathrm{O}_3$ represent zero matrix of order 3 and $\mathrm{I}_3$ is an identity matrix of order 3 . If $\alpha \mathrm{A}+\beta \mathrm{A}^{-1}=4 \mathrm{I}$, then $\alpha+\beta$ is equal to
13
7
12
8
Solution
$\begin{array}{ll}
& \text {Given that }(A-3 I)(A-5 I)=0 \\
\therefore & A^2-3 A-5 A+15 I=0 \\
\therefore & A^2-8 A+15 I=0 \\
\therefore & A^2+15 I=8 A
\end{array}$ Multiplying entire equation by $\frac{A^{-1}}{2}$, we get
$\frac{1}{2} A+\frac{15}{2} A^{-1}=4 I$ Comparing with $\alpha \mathrm{A}+\beta \mathrm{A}^{-1}=4 \mathrm{I}$, we get
$\begin{aligned}
& \alpha=\frac{1}{2} \text { and } \beta=\frac{15}{2} \\
\therefore \quad & \alpha+\beta=\frac{1}{2}+\frac{15}{2}=8
\end{aligned}$