Question

Let A be a \( 3 \times 3 \) matrix with real entries such that \(\det(A) = 6\) and \(\text{trace}(A) = 0\). If \(\det(A + I) = 0\), where I is the \( 3 \times 3 \) identity matrix, then the eigenvalues of A are:

• A. \( -1, 2, 3 \)
• B. \( -1, 2, -3 \)
• C. \( 1, 2, -3 \)
• D. \( -1, -2, 3 \)

Hint:

To solve multiple choice questions on matrix eigenvalues quickly, verify the options directly against the properties: - Sum of values must equal trace. - Product of values must equal determinant. Checking Option (B): \((-1) + 2 + (-3) = 0\) and \((-1) \times 2 \times (-3) = 6\). It satisfies both instantly!

Answer 1

The Correct Option is B

Understanding the Concept: Let the three eigenvalues of the \( 3 \times 3 \) matrix \( A \) be denoted by \( \lambda_1, \lambda_2, \) and \( \lambda_3 \). We use the following essential properties of matrix eigenvalues:
• Trace Property: The sum of the eigenvalues is equal to the trace of the matrix: \[ \sum_{i=1}^{n} \lambda_i = \text{trace}(A) \]
• Determinant Property: The product of the eigenvalues is equal to the determinant of the matrix: \[ \prod_{i=1}^{n} \lambda_i = \det(A) \]
• Shifted Matrix Property: If \(\lambda\) is an eigenvalue of \( A \), then \(\lambda + k\) is an eigenvalue of the shifted matrix \( A + kI \).

Step 1: Translate given matrix conditions into eigenvalue equations
We are given three explicit conditions:
• Trace condition: \[ \lambda_1 + \lambda_2 + \lambda_3 = 0 \quad \cdots (1) \]
• Determinant condition: \[ \lambda_1 \cdot \lambda_2 \cdot \lambda_3 = 6 \quad \cdots (2) \]
• Shifted determinant condition: We know that \(\det(A + I) = 0\). The eigenvalues of the matrix \( A + I \) are \( (\lambda_1 + 1), (\lambda_2 + 1), \) and \( (\lambda_3 + 1) \). The determinant of \( A + I \) is the product of its eigenvalues: \[ (\lambda_1 + 1)(\lambda_2 + 1)(\lambda_3 + 1) = 0 \] For this product to equal zero, at least one of the factor terms must be zero. Let us assume without loss of generality that: \[ \lambda_1 + 1 = 0 \implies \lambda_1 = -1 \]

Step 2: Solve the system of equations for the remaining eigenvalues
Substitute \( \lambda_1 = -1 \) into Equation (1) and Equation (2): From Equation (1): \[ -1 + \lambda_2 + \lambda_3 = 0 \implies \lambda_2 + \lambda_3 = 1 \implies \lambda_3 = 1 - \lambda_2 \quad \cdots (3) \] From Equation (2): \[ (-1) \cdot \lambda_2 \cdot \lambda_3 = 6 \implies \lambda_2 \cdot \lambda_3 = -6 \quad \cdots (4) \] Substitute Equation (3) into Equation (4): \[ \lambda_2(1 - \lambda_2) = -6 \] \[ \lambda_2 - \lambda_2^2 = -6 \] Rearranging into standard quadratic equation form: \[ \lambda_2^2 - \lambda_2 - 6 = 0 \] Factoring the quadratic equation: \[ (\lambda_2 - 3)(\lambda_2 + 2) = 0 \] This gives two possible values for \( \lambda_2 \): \[ \lambda_2 = 3 \quad \text{or} \quad \lambda_2 = -2 \] If \( \lambda_2 = 3 \), then \( \lambda_3 = 1 - 3 = -2 \). If \( \lambda_2 = -2 \), then \( \lambda_3 = 1 - (-2) = 3 \). Thus, the set of eigenvalues of the matrix \( A \) is precisely \(\{-1, 2, -3\}\). This matches Option (B).