Question

Let $A=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3\end{array}\right) .$ Then the roots of the equation $\operatorname{det}\left(A-\lambda I_{3}\right)=0$ (where $I_{3}$ is the identity matrix of order 3 ) are
(A) $3,0,3$
(B) $0,3,6$
(C) $1,0,-6$
(D) $3,3,6$

Answer 1

Ans: (B)
$$\text { Hint : Let } \begin{aligned} \left(A-\lambda I_{3}\right)=0 \quad \Rightarrow\left|\begin{array}{ccc} 3-\lambda & 0 & 3 \\ 0 & 3-\lambda & 0 \\ 3 & 0 & 3-\lambda \end{array}\right|=0 \\ \Rightarrow(3-\lambda)^{3}-9(3-\lambda)=0 \Rightarrow(3-\lambda)\left[(3-\lambda)^{2}-3^{2}\right]=0 \\ \Rightarrow 3-\lambda=0 \text { or } 3-\lambda-3=0 \text { or } 3-\lambda+3=0 \quad \Rightarrow \lambda=0,3 \text { or } 6 \end{aligned}$$

Answer 2

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Answer 3

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