Question

If \[ A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}, \] \(\text{then prove that}\) \[ A \cdot \text{adj} \, A = |A| \cdot I. \]

Hint:

For a matrix \( A \), the product of \( A \) and its adjugate \( \text{adj} \, A \) is always equal to the determinant of \( A \) multiplied by the identity matrix: \( A \cdot \text{adj} \, A = |A| \cdot I \).

Answer 1

The adjugate (adjoint) of a matrix \( A \), denoted by \( \text{adj} \, A \), is the transpose of the cofactor matrix of \( A \). A well-known result in matrix theory is that: \[ A \cdot \text{adj} \, A = |A| \cdot I. \] To prove this, we first calculate \( |A| \), the determinant of \( A \). The determinant of a 3x3 matrix is given by: \[ |A| = \begin{vmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{vmatrix}. \] Using cofactor expansion along the first row: \[ |A| = 1 \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} - 3 \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} + 3 \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix}. \] Now, calculate the 2x2 determinants: \[ \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} = (4 \cdot 4) - (3 \cdot 3) = 16 - 9 = 7, \] \[ \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = (1 \cdot 4) - (3 \cdot 1) = 4 - 3 = 1, \] \[ \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} = (1 \cdot 3) - (4 \cdot 1) = 3 - 4 = -1. \] Substitute these into the determinant formula: \[ |A| = 1 \cdot 7 - 3 \cdot 1 + 3 \cdot (-1) = 7 - 3 - 3 = 1. \] Thus, \( |A| = 1 \). Since the determinant is 1, we have: \[ A \cdot \text{adj} \, A = 1 \cdot I = I. \]
Conclusion: We have proved that \[ A \cdot \text{adj} \, A = |A| \cdot I = I. \]