Question

If \( A = \begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} \) is a \( 2 \times 2 \) matrix, then the eigenvalues of the matrix \( 2A^2 - 4A + 5I \) are ........., where \( I \) is the \( 2 \times 2 \) unit matrix.

• A. $2 \pm i$
• B. $3 \pm 4i$
• C. $3 \pm 2i$
• D. $-3 \pm i$

Hint:

To find eigenvalues, use the characteristic equation $\text{det}(A - \lambda I) = 0$. For matrices with complex eigenvalues, look for negative signs under the square root in the discriminant.

Answer 1

The Correct Option is B

Given the matrix \( A = \begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} \), we need to find the eigenvalues of the matrix \( 2A^2 - 4A + 5I \).

1. First, we calculate \( A^2 \):
\[ A^2 = \begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 7 & -4 \\ 12 & 7 \end{pmatrix} \] 2. Now, calculate \( 2A^2 \):
\[ 2A^2 = 2 \begin{pmatrix} 7 & -4 \\ 12 & 7 \end{pmatrix} = \begin{pmatrix} 14 & -8 \\ 24 & 14 \end{pmatrix} \] 3. Next, calculate \( 4A \):
\[ 4A = 4 \begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 8 & -4 \\ 12 & 8 \end{pmatrix} \] 4. Now calculate \( 5I \) (where \( I \) is the identity matrix):
\[ 5I = 5 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \] 5. Compute \( 2A^2 - 4A + 5I \):
\[ 2A^2 - 4A + 5I = \begin{pmatrix} 14 & -8 \\ 24 & 14 \end{pmatrix} - \begin{pmatrix} 8 & -4 \\ 12 & 8 \end{pmatrix} + \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \] \[ = \begin{pmatrix} 14 - 8 + 5 & -8 + 4 + 0 \\ 24 - 12 + 0 & 14 - 8 + 5 \end{pmatrix} = \begin{pmatrix} 11 & -4 \\ 12 & 11 \end{pmatrix} \] 6. Find the eigenvalues of \( \begin{pmatrix} 11 & -4 \\ 12 & 11 \end{pmatrix} \):
\[ \text{det}\left( \begin{pmatrix} 11 - \lambda & -4 \\ 12 & 11 - \lambda \end{pmatrix} \right) = 0 \] \[ (11 - \lambda)^2 + 48 = 0 \] \[ (11 - \lambda)^2 = -48 \] \[ 11 - \lambda = \pm 4i \Rightarrow \lambda = 11 \mp 4i \] Thus, the eigenvalues are: \( \boxed{11 \pm 4i} \)