Question

Let \( A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \) be a 3 × 3 matrix. If \( \alpha \) and \( \beta \) are the largest and smallest eigenvalues of \( A \), respectively, then \( \alpha - \beta = \_\_\_\_\_\_ \)

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For finding eigenvalues, use the characteristic equation and make sure to calculate the roots carefully.

Answer 1

The Correct Option is D

To solve for the difference between the largest and smallest eigenvalues of matrix \( A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \), we need to find its eigenvalues. The eigenvalues \( \lambda \) are solutions to the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix.

First, let's compute \( A - \lambda I \):

\( \begin{pmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \end{pmatrix} \).

The determinant of this matrix is given by:

\( \det(A - \lambda I) = -\lambda(-\lambda^2 + 1 + 1) - 1(-\lambda + 1) + 1(-\lambda + 1) \).

Simplifying, we have:

\( -\lambda(-\lambda^2 + 2) + 2 \).

Therefore, the characteristic polynomial is:

\( \lambda^3 - 2\lambda - 2 = 0 \).

Using the rational root theorem and testing possible roots, we find that \( \lambda = 2 \) is a root. We perform polynomial division to factor the cubic polynomial:

\( (\lambda - 2)(\lambda^2 + 2\lambda + 1) = 0 \).

The remaining quadratic \( \lambda^2 + 2\lambda + 1 = 0 \) factors further as:

\( (\lambda + 1)^2 = 0 \).

Thus, the eigenvalues are \(\lambda = 2\) and \(\lambda = -1\), with \(\lambda = -1\) having multiplicity 2.

Here, \(\alpha = 2\) and \(\beta = -1\). Thus, \(\alpha - \beta = 2 - (-1) = 3\).

Therefore, the answer is 3.