Question

The eigenvalues of the matrix \[ \begin{bmatrix} 1 & 2
0 & 3 \end{bmatrix} \] are \_\_\_\_\_ .

• A. 0, 2
• B. 2, 3
• C. 1, 3
• D. 1, 2

Hint:

To find the eigenvalues of a matrix, solve the characteristic equation \( {det}(A - \lambda I) = 0 \). The roots of the equation are the eigenvalues.

Answer 1

The Correct Option is C

To find the eigenvalues of a matrix, we solve the characteristic equation: \[ {det}(A - \lambda I) = 0 \] where \( A \) is the matrix, \( \lambda \) is the eigenvalue, and \( I \) is the identity matrix. Step 1: Construct the matrix \( A - \lambda I \)
Given the matrix \( A = \begin{bmatrix} 1 & 2
0 & 3 \end{bmatrix} \), we subtract \( \lambda I \) (where \( I \) is the 2x2 identity matrix) from \( A \): \[ A - \lambda I = \begin{bmatrix} 1 & 2
0 & 3 \end{bmatrix} - \begin{bmatrix} \lambda & 0
0 & \lambda \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 2
0 & 3 - \lambda \end{bmatrix} \] Step 2: Find the determinant of \( A - \lambda I \)
Now, compute the determinant of the resulting matrix: \[ {det}(A - \lambda I) = {det}\begin{bmatrix} 1 - \lambda & 2
0 & 3 - \lambda \end{bmatrix} \] The determinant of a 2x2 matrix \( \begin{bmatrix} a & b
c & d \end{bmatrix} \) is given by \( ad - bc \). So, for our matrix: \[ {det}(A - \lambda I) = (1 - \lambda)(3 - \lambda) - (0)(2) = (1 - \lambda)(3 - \lambda) \] Step 3: Solve the characteristic equation
Now, we solve the characteristic equation: \[ (1 - \lambda)(3 - \lambda) = 0 \] This gives two solutions: \[ 1 - \lambda = 0 \quad \Rightarrow \quad \lambda = 1 \] \[ 3 - \lambda = 0 \quad \Rightarrow \quad \lambda = 3 \] Conclusion: Thus, the eigenvalues of the matrix are \( \lambda = 1 \) and \( \lambda = 3 \).