Question

An Eigen value of the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$ is $1$. An eigen vector corresponding to it is:}

• A. $\begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix}$
• B. $\begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix}$
• C. $\begin{bmatrix} -4 \\ 2
  1 \end{bmatrix}$
• D. $\begin{bmatrix} 3 \\ 0
  1 \end{bmatrix}$

Hint:

To quickly verify an option vector as an eigenvector in competitive examinations, multiply the matrix $A$ directly by the choice vectors. Whichever choice yields a scalar multiple of itself ($AX = 1 \cdot X$) is the correct response.

Answer 1

The Correct Option is C

Concept:& nbsp;
By definition, a non-zero vector \(X\) is an eigenvector of a square matrix \(A\) corresponding to an eigenvalue \(\lambda\) if it satisfies \[ AX=\lambda X \] or equivalently, \[ (A-\lambda I)X=O, \] where \(I\) is the identity matrix and \(O\) is the zero column vector. Given the eigenvalue \(\lambda=1\), we determine the eigenvectors by solving \[ (A-I)X=O. \]& nbsp;

Step 1: Construct the matrix \((A-I)\).
Let \[ A= \begin{bmatrix} 1 & -1 & 2\\ 0 & 1 & 0\\ 1 & 2 & 1 \end{bmatrix}. \] Then \[ A-I= \begin{bmatrix} 1-1 & -1 & 2\\ 0 & 1-1 & 0\\ 1 & 2 & 1-1 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 2\\ 0 & 0 & 0\\ 1 & 2 & 0 \end{bmatrix}. \]& nbsp;

Step 2: Set up the homogeneous system.
Let \[ X= \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}. \] Then \[ \begin{bmatrix} 0 & -1 & 2\\ 0 & 0 & 0\\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}. \] This gives the equations: \[ -x_2+2x_3=0 \] \[ x_1+2x_2=0. \]& nbsp;

Step 3: Solve the system.
From \[ -x_2+2x_3=0, \] we get \[ x_2=2x_3. \] Substituting into \[ x_1+2x_2=0, \] gives \[ x_1=-2x_2=-4x_3. \] Let \[ x_3=k, \] where \(k\neq 0\). Then \[ X= \begin{bmatrix} -4k\\ 2k\\ k \end{bmatrix} = k \begin{bmatrix} -4\\ 2\\ 1 \end{bmatrix}. \]& nbsp;

Step 4: Identify the correct option.
Taking \(k=1\), \[ X= \begin{bmatrix} -4\\ 2\\ 1 \end{bmatrix}. \] Therefore, the eigenvector corresponding to \(\lambda=1\) is \[ \boxed{ \begin{bmatrix} -4\\ 2\\ 1 \end{bmatrix} } \] Hence, the correct answer is \[ \boxed{\text{Option (C)}}. \]