Question

The eigen vectors of the matrix \(\begin{bmatrix} 1 & 2 \\ 0 & 4 \end{bmatrix}\) are in the form \(\begin{bmatrix} 1 \\ a \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ b \end{bmatrix}\), then \( a + b =\)

• A. 0
• B. 1
• C. \(\frac{3}{2}\)
• D. \(\frac{2}{3}\)

Hint:

To find eigenvectors, solve \( (A - \lambda I) \mathbf{v} = 0 \) for each eigenvalue \( \lambda \). The resulting vectors can be scaled to match the required form.

Answer 1

The Correct Option is C

Step 1: Find the eigenvalues of the matrix.
Given the matrix \( A = \begin{bmatrix} 1 & 2 \\ 0 & 4 \end{bmatrix} \), we first compute the eigenvalues by solving the characteristic equation: \[ \det(A - \lambda I) = 0, \] where \( I \) is the identity matrix and \( \lambda \) is an eigenvalue. The matrix \( A - \lambda I \) is: \[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 2 \\ 0 & 4 - \lambda \end{bmatrix}. \] The determinant is: \[ \det(A - \lambda I) = (1 - \lambda)(4 - \lambda) - (2)(0) = (1 - \lambda)(4 - \lambda). \] Set the determinant to zero: \[ (1 - \lambda)(4 - \lambda) = 0, \] \[ \lambda = 1 \quad \text{or} \quad \lambda = 4. \] So, the eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 4 \).
Step 2: Find the eigenvector for \( \lambda = 1 \).
For \( \lambda = 1 \), compute \( A - \lambda I \): \[ A - 1 \cdot I = \begin{bmatrix} 1 - 1 & 2 \\ 0 & 4 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 0 & 3 \end{bmatrix}. \] Solve the system \( (A - \lambda I) \mathbf{v} = 0 \), where \( \mathbf{v} = \begin{bmatrix} x \\ y \end{bmatrix} \): \[ \begin{bmatrix} 0 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \] This gives the equations: \[ 0x + 2y = 0 \quad \Rightarrow \quad 2y = 0 \quad \Rightarrow \quad y = 0, \] \[ 0x + 3y = 0 \quad \Rightarrow \quad 3y = 0 \quad \Rightarrow \quad y = 0. \] The first equation confirms \( y = 0 \), and \( x \) is free. Let \( x = t \), so the eigenvector is: \[ \mathbf{v}_1 = t \begin{bmatrix} 1 \\ 0 \end{bmatrix}. \] The problem states the eigenvector is of the form \( \begin{bmatrix} 1 \\ a \end{bmatrix} \), so set \( t = 1 \): \[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \] thus \( a = 0 \).

Step 3: Find the eigenvector for \( \lambda = 4 \).
For \( \lambda = 4 \), compute \( A - \lambda I \): \[ A - 4 \cdot I = \begin{bmatrix} 1 - 4 & 2 \\ 0 & 4 - 4 \end{bmatrix} = \begin{bmatrix} -3 & 2 \\ 0 & 0 \end{bmatrix}. \] Solve the system \( (A - \lambda I) \mathbf{v} = 0 \), where \( \mathbf{v} = \begin{bmatrix} x \\ y \end{bmatrix} \): \[ \begin{bmatrix} -3 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \] This gives the equations: \[ -3x + 2y = 0 \quad \Rightarrow \quad 2y = 3x \quad \Rightarrow \quad y = \frac{3}{2}x, \] \[ 0x + 0y = 0 \quad \Rightarrow \quad \text{trivial equation}. \] Let \( x = t \), then \( y = \frac{3}{2}t \), so the eigenvector is: \[ \mathbf{v}_2 = t \begin{bmatrix} 1 \\ \frac{3}{2} \end{bmatrix}. \] The problem states the eigenvector is of the form \( \begin{bmatrix} 1 \\ b \end{bmatrix} \), so set \( t = 1 \): \[ \mathbf{v}_2 = \begin{bmatrix} 1 \\ \frac{3}{2} \end{bmatrix}, \] thus \( b = \frac{3}{2} \).

Step 4: Compute \( a + b \).
From the eigenvectors: - \( a = 0 \) (for \( \lambda = 1 \)), - \( b = \frac{3}{2} \) (for \( \lambda = 4 \)). \[ a + b = 0 + \frac{3}{2} = \frac{3}{2}. \]

Step 5: Evaluate the options.
(1) 0: Incorrect, as \( a + b = \frac{3}{2} \). Incorrect.
(2) 1: Incorrect, as \( a + b = \frac{3}{2} \). Incorrect.
(3) \(\frac{3}{2}\): Correct, as \( a + b = \frac{3}{2} \). Correct.
(4) \(\frac{2}{3}\): Incorrect, as \( a + b = \frac{3}{2} \). Incorrect.

Step 6: Select the correct answer. The sum \( a + b = \frac{3}{2} \), matching option (C).