If $\mathrm{A}$ is non-singular matrix of order 3 such that $(A-2 \mathrm{I})(A-4 \mathrm{I})=0$, then $\frac{1}{6} A+\frac{4}{3} A^{-1}$ is (where I is a unit matrix of order 3 and 0 is a null matrix of order 3)

Question

Aicharya · Accepted Answer

$\begin{aligned} & (A-2 \mathrm{I})(A-4 \mathrm{I})=0 \ & \Rightarrow A^2-6 A+8 \mathrm{I}=0 \ & \Rightarrow A^2+8 \mathrm{I}=6 A \ & \Rightarrow A^2 A^{-1}+8 \mathrm{I} A^{-1}=6 A A^{-1} \ & \Rightarrow A+8 A^{-1}=6 \mathrm{I} \ & \Rightarrow \frac{1}{6} A+\frac{4}{3} A^{-1}=\mathrm{I}\end{aligned}$

If $\mathrm{A}$ is non-singular matrix of order 3 such that $(A-2 \mathrm{I})(A-4 \mathrm{I})=0$, then…

Solution