Question

Let \( A \) be a \( 3 \times 3 \) matrix such that \( X^T AX = 0 \) for all nonzero \( 3 \times 1 \) matrices \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \). \[ A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix} \] If \( \det(\text{adj}(2(A + I))) = 2^{\alpha} 3^{\beta} 5^{\gamma} \), \( \alpha, \beta, \gamma \in \mathbb{N} \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is:

Hint:

When working with determinants of matrices, be sure to use the properties of the adjugate matrix and the fact that the determinant of the adjugate is related to the determinant of the original matrix. Pay attention to the exponentiation for power terms when dealing with matrices.

Answer 1

Step 1: Understand the condition \( X^T A X = 0 \). If \( X^T A X = 0 \) for all nonzero vectors \( X \), then \( A \) must be a skew-symmetric matrix. Therefore, \( A^T = -A \).

Step 2: Use the given information to find \( A \). Let \( v_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \) and \( v_2 = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \). We are given:

\[ A v_1 = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix} \quad \text{and} \quad A v_2 = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix}. \] Also, let \( v_3 = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) be an arbitrary vector. Since \( A \) is skew-symmetric, the diagonal entries must be zero.

Let us compute \( A v_1 + A v_2 \):

\[ A v_1 + A v_2 = A(v_1 + v_2) = \begin{bmatrix} 1 \\ 8 \\ -13 \end{bmatrix}. \] Since we do not know enough about \( A \), let us proceed. The relation \( X^T A X = 0 \) for all \( X \) implies \( A \) is skew-symmetric. Let \( v_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, v_2 = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \), then: \[ A v_1 = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}, \quad A v_2 = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix}. \] From \( A^T = -A \), we know \( a_{ii} = 0 \) for all \( i \). Let \( A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \). Then: \[ A v_1 = \begin{bmatrix} a+b \\ -a+c \\ -b-c \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix} \] and \[ A v_2 = \begin{bmatrix} 2a+b \\ -a+2c \\ -2b-c \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix}. \] So we have the following system of equations: \[ a + b = 1, \quad -a + c = 4, \quad -b - c = -5, \] \[ 2a + b = 0, \quad -a + 2c = 4, \quad -2b - c = -8. \] From \( a + b = 1 \) and \( 2a + b = 0 \), we get \( a = -1 \) and \( b = 2 \). From \( -a + c = 4 \), we get \( c = 3 \). Therefore, the matrix \( A \) is: \[ A = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}. \]

Step 3: Calculate \( \det(2(A + I)) \)

We first compute \( A + I \), where \( I \) is the identity matrix: \[ A + I = \begin{bmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{bmatrix}. \] Then: \[ 2(A + I) = \begin{bmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 \end{bmatrix}. \] The determinant of \( 2(A + I) \) is calculated as: \[ \det(2(A + I)) = 2(4 + 36) + 2(4 + 24) + 4(-12 + 8) = 80 + 56 - 16 = 120. \]

Step 4: Calculate \( \det(\text{adj}(2(A + I))) \)

We know that: \[ \det(\text{adj}(M)) = (\det M)^{n-1}, \] where \( n \) is the size of the matrix. In our case, \( n = 3 \), so: \[ \det(\text{adj}(2(A + I))) = (\det(2(A + I)))^{3-1} = (120)^2 = 2^6 \cdot 3^2 \cdot 5^2. \]

Step 5: Find \( \alpha, \beta, \gamma \) and \( \alpha^2 + \beta^2 + \gamma^2 \)

We have: \[ \det(\text{adj}(2(A + I))) = 2^6 \cdot 3^2 \cdot 5^2, \] so \( \alpha = 6 \), \( \beta = 2 \), and \( \gamma = 2 \). Thus: \[ \alpha^2 + \beta^2 + \gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44. \]

Final Answer:

The value of \( \alpha^2 + \beta^2 + \gamma^2 \) is \( \boxed{44} \).