Question

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

Hint:

For any square matrix \( A \), the relation \( A \cdot \text{adj}(A) = |A| \cdot I \) holds true, and the inverse of \( A \) is given by \( A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) \).

Answer 1

We need to prove that \( A \cdot \text{adj}(A) = |A| \cdot I \), and also find \( A^{-1} \).
Step 1: Find the Determinant of \( A \) The determinant of matrix \( A \) is given by: \[ |A| = \begin{vmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{vmatrix}. \] Expanding 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, compute the 2x2 determinants: \[ \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} = (4 \times 4) - (3 \times 3) = 16 - 9 = 7, \] \[ \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = (1 \times 4) - (1 \times 3) = 4 - 3 = 1, \] \[ \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} = (1 \times 3) - (1 \times 4) = 3 - 4 = -1. \] Substitute these into the determinant formula: \[ |A| = 1 \times 7 - 3 \times 1 + 3 \times (-1) = 7 - 3 - 3 = 1. \] Thus, \( |A| = 1 \).
Step 2: Verify the Formula \( A \cdot \text{adj}(A) = |A| \cdot I \) The adjugate (adjoint) of \( A \), denoted by \( \text{adj}(A) \), is the transpose of the cofactor matrix of \( A \). Since \( |A| = 1 \), we have: \[ A \cdot \text{adj}(A) = |A| \cdot I = 1 \cdot I = I. \] Thus, \( A \cdot \text{adj}(A) = I \), proving the required result.
Step 3: Find the Inverse of \( A \) The inverse of \( A \) is given by: \[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A). \] Since \( |A| = 1 \), we have: \[ A^{-1} = \text{adj}(A). \] So, we need to find the adjugate matrix \( \text{adj}(A) \), which is the transpose of the cofactor matrix of \( A \). The cofactor matrix is calculated by finding the cofactor of each element in \( A \), and then taking the transpose. The cofactor matrix \( C(A) \) is: \[ C(A) = \begin{bmatrix} 7 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 7 \end{bmatrix}. \] The adjugate matrix \( \text{adj}(A) \) is the transpose of \( C(A) \): \[ \text{adj}(A) = \begin{bmatrix} 7 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 7 \end{bmatrix}. \] Thus, the inverse of \( A \) is: \[ A^{-1} = \begin{bmatrix} 7 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 7 \end{bmatrix}. \]
Conclusion: We have proved that \[ A \cdot \text{adj}(A) = I, \] and the inverse of \( A \) is \[ A^{-1} = \begin{bmatrix} 7 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 7 \end{bmatrix}. \]