Question

If $A$ is nonsingular and $(A-2I)(A-4I)=0$, then $\frac{1}{6}A+\frac{4}{3}{A}^{-1}$ is equal to

• A. $0$
• B. $I$
• C. $2I$
• D. $6I$

Answer 1

The correct option is B $I$

Since, A is non-singular
So, $A^{-1}$ exists.
Given, $(A-2I)(A-4I)=0$
$\Rightarrow A^2-6A+8I=0$
Multiplying by $A^{-1}$, we get
$\Rightarrow A-6I+8A^{-1}=0$
$\Rightarrow A+8A^{-1}=6I$
$\Rightarrow \frac{1}{6}A+\frac{4}{3}{A}^{-1}=I$