Question

Let $M =\begin{bmatrix} \alpha & − 6 \\-1 & -1 \end{bmatrix}, \alpha\, \epsilon\, R $ be a 2 × 2 matrix. If the eigenvalues of $M$ are $\beta$ and 4, then which of the following is/are CORRECT?

• A. $\alpha + \beta = 1$
• B. An eigenvector corresponding to $\beta$ is $[2, 1]^T$
• C. The rank of the matrix M is 2
• D. The matrix $M^2$ + M is invertible

Answer 1

The Correct Option is A, B, C

Eigenvalue and Matrix Analysis 

Given Matrix:

The matrix M is:

M = | α 1 |
| -6 1 |

Its eigenvalues are β and 4.

Step 1: Trace of the Matrix

The trace of a matrix is the sum of its diagonal elements and equals the sum of its eigenvalues:

Trace(M) = α + 1

Since the eigenvalues are β and 4:

Trace(M) = β + 4

Equating these:

α + 1 = β + 4 ⇒ α + β = 1

Conclusion: Option (A) is correct.

Step 2: Eigenvector Corresponding to β

To find the eigenvector corresponding to β, we solve:

(M − βI) v = 0, where v is the eigenvector.

Subtracting βI from M:

M − βI = | α − β 1 |
| -6 1 − β |

The eigenvector v satisfies:

| α − β 1 | |x| = 0
| -6 1 − β | |y|

Assume v = | 2 | 1 |. Substituting:

• 2(α − β) + 1 = 0
• -12 + 1 − β = 0

Solving -12 + 1 − β = 0 gives β = -11.

Conclusion: Option (B) is correct.

Step 3: Rank of the Matrix M

The rank of M is the number of linearly independent rows or columns.

Since M has distinct eigenvalues (β and 4), the rank of M is 2, meaning it is invertible.

Conclusion: Option (C) is correct.

Step 4: Invertibility of M² + M

To check if M² + M is invertible, consider:

M² + M = M (M + I)

For M² + M to be invertible, neither M nor M + I should have zero as an eigenvalue.

However, further computation of the eigenvalues of M + I reveal that it may not satisfy this condition for all α.

Conclusion: Option (D) is incorrect.

Final Answer:

The correct options are:

• (A) α + β = 1
• (B) β = -11 satisfies the eigenvector equation.
• (C) The rank of M is 2.