Question

Consider the matrix:

\[ A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \]

The eigenvalues of the matrix are 0.27 and ____ (rounded off to 2 decimal places).

Hint:

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

Answer 1

The eigenvalues of a matrix \( A \) are found by solving the characteristic equation:

\[ \det(A - \lambda I) = 0 \]

Where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix.

For the given matrix \( A \):

\[ A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \quad \Rightarrow \quad A - \lambda I = \begin{bmatrix} 2 - \lambda & 3 \\ 1 & 2 - \lambda \end{bmatrix} \]

Now, calculate the determinant:

\[ \det(A - \lambda I) = (2 - \lambda)(2 - \lambda) - 3 \cdot 1 \] \[ = (2 - \lambda)^2 - 3 \] \[ = 4 - 4\lambda + \lambda^2 - 3 = \lambda^2 - 4\lambda + 1 \]

Set the determinant equal to zero to find the eigenvalues:

\[ \lambda^2 - 4\lambda + 1 = 0 \]

Solve this using the quadratic formula:

\[ \lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 4}}{2} \] \[ \lambda = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 3.464}{2} \]

Thus, the two eigenvalues are:

\[ \lambda_1 = \frac{4 + 3.464}{2} = 3.73, \quad \lambda_2 = \frac{4 - 3.464}{2} = 0.27 \]

Therefore, the second eigenvalue is 3.73.