Question

Matrix $A$ satisfies $A^2=3A-2I.$ If $A^{-1}=\frac{\lambda I+k{A}^3}{\mu },$ then the value of $\lambda +k\mu$, where $I$ is the identity matrix of order same as matrix $A$, is

• A. $7$
• B. $1$
• C. $-2$
• D. $-3$

Answer 1

The correct option is B $1$

$A^2=3A-2I$
Pre-multiplying both sides by $A$, we get
$A^3=3A^2-2A$
$\Rightarrow A^3=3(3A-2I)-2A$
$\Rightarrow A^3=7A-6I$
$A^4=7A^2-6A=7(3A-2I)-6A$
$\Rightarrow A^4=15A-14I$
Pre-multiplying both sides by $A^{-1}$, we get
$A^3=15I-14A^{-1}$
or, $A^{-1}=\frac{15I-A^3}{14}$

Comparing with $A^{-1}=\frac{\lambda I+k{A}^3}{\mu }$, we get
$\lambda =15,k=-1,\mu =14$
$\therefore \lambda +ku=15+(-1)14=15-14=1$