Q111 / 155 · WBJEE 2019 · Mathematics

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
  • Correct answer
    B
    0,3,6
  • C
    1,0,-6
  • D
    3,3,6

Solution

We have, $$ A=\left[\begin{array}{lll} 3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3 \end{array}\right] $$ Again, we have $$ \left|\begin{array}{ccc} 3-\lambda & 0 & 3 \\ 0 & 3-\lambda & 0 \\ 3 & 0 & 3-\lambda \end{array}\right|=0 $$ $\Rightarrow(3-\lambda)\left[(3-\lambda)^{2}-0\right]-0+3[0-3(3-\lambda)]=0$ $\Rightarrow (3-\lambda)^{3}-9(3-\lambda)=0$ $\Rightarrow (3-\lambda)\left[(3-\lambda)^{2}-9\right]=0$ $\Rightarrow (3-\lambda)\left(9+\lambda^{2}-6 \lambda-9\right)=0$ $\Rightarrow (3-\lambda)\left(\lambda^{2}-6 \lambda\right)=0$ $\Rightarrow \quad(3-\lambda) \lambda(\lambda-6)=0$ ${\lambda}=3,0,6$

Hence, the correct answer is option: 'B'.

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