Question

The characteristic equation of a matrix $A$ is ${\lambda }^3-5{\lambda }^2-3\lambda +2I=0$ then $|adjA|=$

• A. $4$
• B. $25$
• C. $9$
• D. $30$

Answer 1

The correct option is A $4$

Since characteristic equation of degree 3 so $A$ is a $3\times 3$ matrix

Charectristic equation =$P\left(\lambda \right)=|A-\lambda I|={\lambda }^3-5{\lambda }^2-3\lambda +2I$

So $|A|=P\left(0\right)=2$

So $A^{-1}=\frac{1}{|A|}adjA$

$|A^{-1}|=|\frac{1}{|A|}adjA|=(\frac{1}{|A|}{)}^3|adjA|$

Since $|A^{-1}|=|A{|}^{-1}$

$⟹|A{|}^2=|adjA|=4$

Hence the answer