Question

The eigen vectors of the matrix $A=\begin{bmatrix}5 & 4\\ 1& 2\end{bmatrix}$ is

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

Hint:

To save time, check the options using $AX = \lambda X$. Multiply the matrix $A$ by the given vectors. If the result is a scalar multiple of the vector, it's an eigenvector! For (3): $\begin{bmatrix} 5 & 4
1 & 2 \end{bmatrix} \begin{bmatrix} 4
1 \end{bmatrix} = \begin{bmatrix} 24
6 \end{bmatrix} = 6 \begin{bmatrix} 4
1 \end{bmatrix}$. Verified!

Answer 1

The Correct Option is C

Solution: Finding Eigenvalues and Eigenvectors

Step 1: Find the Characteristic Equation and Eigenvalues
The characteristic equation is given by:
$$|A - \lambda I| = \begin{vmatrix} 5-\lambda & 4 \\ 1 & 2-\lambda \end{vmatrix} = 0$$
Expanding the determinant:
$$(5-\lambda)(2-\lambda) - (4)(1) = 0$$
$$\lambda^2 - 7\lambda + 6 = 0$$
Factoring the quadratic:
$$(\lambda - 6)(\lambda - 1) = 0$$
Therefore, the eigenvalues are:
$$\lambda_1 = 6, \quad \lambda_2 = 1$$

Step 2: Find Eigenvector for $\lambda = 6$
Substitute $\lambda = 6$ into $(A - \lambda I)X = 0$:
$$\begin{bmatrix} 5-6 & 4 \\ 1 & 2-6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} -1 & 4 \\ 1 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
This gives the equation:
$$-x + 4y = 0 \implies x = 4y$$
Let $y = 1$, then $x = 4$.
Eigenvector: $$v_1 = \begin{bmatrix} 4 \\ 1 \end{bmatrix}$$

Step 3: Find Eigenvector for $\lambda = 1$
Substitute $\lambda = 1$ into $(A - \lambda I)X = 0$:
$$\begin{bmatrix} 5-1 & 4 \\ 1 & 2-1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} 4 & 4 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
This gives the equation:
$$x + y = 0 \implies x = -y$$
Let $y = -1$, then $x = 1$.
Eigenvector: $$v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$

Step 4: Match with Options
The eigenvectors are proportional to:
$$\begin{bmatrix} 4 \\ 1 \end{bmatrix} \text{ and } \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$
This corresponds to Option (3).

Answer: Option (3)