ISI2021-MMA: 21

Question

Suppose that a $3 \times 3$ matrix $A$ has an eigen value $-1$. If the matrix $A+I$ is equal to 

$$ \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $$

then the eigen vectors of $A$ corresponding to the eigenvalue $-1$ are in the form,

• A. $\left[\begin{array}{c}2 t \\ 0 \\ t\end{array}\right], t \in \mathbb{R}$
• B. $\left[\begin{array}{c}2 t \\ s \\ t\end{array}\right], s, t \in \mathbb{R}$
• C. $\left[\begin{array}{c}t \\ 0 \\ -2 t\end{array}\right], t \in \mathbb{R}$
• D. $\left[\begin{array}{c}t \\ s \\ 2 t\end{array}\right], s, t \in \mathbb{R}$

Answer 1

correct answer  D.

-1 is an eigen value of A then from definition we have

AX=(-1)X, where X is a non zero vector.

(A+I)X=0

$\begin{align} \begin{bmatrix} 1 &0 &-2 \\ 0& 0& 0\\ 0&0 &0 \end{bmatrix}X=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \end{align}$

So the vector will be, $\begin{bmatrix} t\\ s\\ 2t \end{bmatrix},t,s\in\mathbb{R}$

Comments

Incorrect

Answer 2

$A+I=\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

$A=\left[\begin{array}{ccc}0 & 0 & -2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$

We know, $AX = \lambda X$ here $\lambda=-1$ given.

$\left[\begin{array}{ccc}0 & 0 & -2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{c} a \\ b \\ c\end{array}\right] = (-1)\left[\begin{array}{c} a \\ b \\ c\end{array}\right]$

$\left[\begin{array}{c} -2c \\ -b \\ -c\end{array}\right]=\left[\begin{array}{c} -a \\ -b \\ -c\end{array}\right]$

$\therefore a=2c, b=b, c=c$

$\color{Green} Ans: \text{B.} \left[\begin{array}{c} 2t \\ s \\ t\end{array}\right],s,t \in \mathbb{R}$