Question

If \( A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} \) 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:

You can apply polynomials directly on eigenvalues. If $\lambda$ is an eigenvalue of $A$, then $f(\lambda)$ is an eigenvalue of $f(A)$.

Answer 1

The Correct Option is C

Step 1: Find eigenvalues of A
The characteristic equation of \( A \) is:
\[ \det(A - \lambda I) = \begin{vmatrix} 1 - \lambda & -1 \\ 2 & 3 - \lambda \end{vmatrix} = (1 - \lambda)(3 - \lambda) - (-1)(2) = \lambda^2 - 4\lambda + 5 = 0 \]
Solving:
\[ \lambda = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5)}}{2} = \frac{4 \pm \sqrt{-4}}{2} = 2 \pm i \]
Step 2: Use eigenvalue transformation rule
If \( \lambda \) is an eigenvalue of \( A \), then \( f(\lambda) \) is an eigenvalue of \( f(A) \) for any polynomial \( f \). Let \( f(x) = 2x^2 - 4x + 5 \):
\[ \text{Let } \lambda = 2 \pm i \Rightarrow f(2 \pm i) = 2(2 \pm i)^2 - 4(2 \pm i) + 5 \]
Now compute:
\[ (2 \pm i)^2 = 4 \pm 4i + i^2 = 3 \pm 4i \]
\[ f(2 \pm i) = 2(3 \pm 4i) - 8 \mp 4i + 5 = 6 \pm 8i - 8 \mp 4i + 5 = 3 \pm 2i \]
So, eigenvalues of \( 2A^2 - 4A + 5I \) are:
\[ \boxed{3 \pm 2i} \]