Question

Let

\[ M = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, \quad K = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} \]

satisfy the eigenvalue problem:

\[ (M - \alpha K)\,\phi = 0 \]

The lowest eigenvalue \( \alpha \) is __________ (rounded off to two decimal places).

Hint:

When solving eigenvalue problems, remember to set the determinant of \( (M - \alpha K) \) to zero. The solutions to the resulting equation give the eigenvalues.

Answer 1

Correct Answer: 0.42

We are given the matrix equation \( (M - \alpha K)\phi = 0 \), which represents an eigenvalue problem.
For a non-trivial solution \( \phi \), the determinant of \( (M - \alpha K) \) must be zero.

Thus, we solve:
\[ \det(M - \alpha K) = 0 \]

First, compute \( M - \alpha K \):
\[ M - \alpha K = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} - \alpha \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 - 2\alpha & \alpha \\ \alpha & 2 - \alpha \end{bmatrix} \]

Now compute the determinant:
\[ \det(M - \alpha K) = (1 - 2\alpha)(2 - \alpha) - \alpha^2 \]

Expanding:
\[ = 2 - \alpha - 4\alpha + 2\alpha^2 - \alpha^2 = 2 - 5\alpha + \alpha^2 \]

Setting the determinant equal to zero:
\[ \alpha^2 - 5\alpha + 2 = 0 \]

Using the quadratic formula:
\[ \alpha = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2} \]

Thus, the two eigenvalues are:
\[ \alpha = \frac{5 + \sqrt{17}}{2} \quad \text{and} \quad \alpha = \frac{5 - \sqrt{17}}{2} \]

The lowest eigenvalue is:
\[ \alpha = \frac{5 - \sqrt{17}}{2} \approx 0.44 \]

Final Answer: 0.44