Question

Let A be a $3\times 3$ matrix with real entries such that $A^2=A+2I$, where I denotes the $3\times 3$ identity matrix. If $\alpha ,\beta and\gamma$ are eigen value of A such that $\alpha \beta \gamma =-4$ than $\alpha +\beta +\gamma$ is equal to

• A. -1
• B. 0
• C. 2
• D. 3

Answer 1

The correct option is D 3

Given $A^2=A+2I$

$A^2-A-2I=0$

By cayley - Hamilton theorem

${\lambda }^2-\lambda -2=0$

Where, value of $\lambda$ are eigen values of A.

$\Rightarrow {\lambda }^2-2\lambda +\lambda -2=0$

$\Rightarrow (\lambda -2)(\lambda +1)=0$

So, ${\lambda }_1=\alpha =2,{\lambda }_2=\beta =-1$

$\because \alpha \cdot \beta \cdot \gamma =-4$

$\Rightarrow$ Third eigen value is $\gamma =\frac{-4}{\alpha \cdot \beta }=\frac{-4}{2\times -1}=2$

So, $\alpha +\beta +\gamma =2-1+2=3$