Question

Let
\(\begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ -2 & 2 & 2 \end{pmatrix}\)
and B = A⁵ + A⁴ + I₃. Which of the following is NOT an eigenvalue of B ?

• A. 1
• B. 2
• C. 49
• D. 3

Answer 1

The Correct Option is B

To determine which eigenvalues are valid for the matrix \( B = A^5 + A^4 + I_3 \), we need to explore the eigenvalues of the given matrix \( A \) and how matrix powers affect these eigenvalues.

The given matrix \( A \) is: 

\[A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ -2 & 2 & 2 \end{pmatrix}\]

First, we need to find the eigenvalues of matrix \( A \). The eigenvalue equation is:

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

where \( I \) is the identity matrix and \( \lambda \) is an eigenvalue of \( A \).

Substituting the values, we have:

\[\text{det}\left(\begin{pmatrix} 1 - \lambda & -1 & 0 \\ 0 & -\lambda & 0 \\ -2 & 2 & 2 - \lambda \end{pmatrix}\right) = 0\]

The determinant of a matrix with a row or column of all zeros is always zero. Here, the second row is all zeros. Hence, a direct computation shows that eigenvalues of \( A \) include 0. Calculating the full determinant by expanding along the second row helps verify:

\[\text{det}(A - \lambda I) = (-1)\cdot [(-1)\cdot((2-\lambda)(0) - (0)(2)) + 0 - 0] = -\lambda (1 - \lambda)(2 - \lambda)\]

This gives the characteristic equation:

\[(-\lambda)((1 - \lambda)(2 - \lambda)) = 0\]

Simplifying, we find the eigenvalues:

• \( \lambda_1 = 0 \)
• \( \lambda_2 = 1 \)
• \( \lambda_3 = 2 \)

Since matrix powers retain the eigenvector directions, each eigenvalue \(\lambda\) of \( A \) is also an eigenvalue of \( A^n \). Thus, the eigenvalues of \( B = A^5 + A^4 + I_3 \) are determined by:

\[\lambda_B = \lambda^5 + \lambda^4 + 1\]

Testing the eigenvalues of \( A \):

• For \(\lambda = 0\):
• For \(\lambda = 1\):
• For \(\lambda = 2\):

Thus, the eigenvalues of \( B \) are 1, 3, and 49. Therefore, the value that is NOT an eigenvalue of \( B \) is:

2