## Determinant of Matrix \( A \) Given: \[ A = \begin{bmatrix} 4 & 3 & 3 & 3 \\ 3 & 4 & 3 & 3 \\ 3 & 3 & 4 & 3 \\ 3 & 3 & 3 & 4 \\ \end{bmatrix} \] This is a symmetric matrix where diagonal elements are 4 and off-diagonal elements are 3. ### Step 1: Row Operations Subtract the last row from all previous rows: - \( R_1' = R_1 - R_4 \) - \( R_2' = R_2 - R_4 \) - \( R_3' = R_3 - R_4 \) \[ \begin{bmatrix} 4-3 & 3-3 & 3-3 & 3-4 \\ 3-3 & 4-3 & 3-3 & 3-4 \\ 3-3 & 3-3 & 4-3 & 3-4 \\ 3 & 3 & 3 & 4 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 3 & 3 & 3 & 4 \\ \end{bmatrix} \] ### Step 2: Expanding the Determinant Let’s expand along the first three rows (which now have many zeros): \[ \text{det}(A) = \begin{vmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 3 & 3 & 3 & 4 \\ \end{vmatrix} \] Expand along the first row: \[ = 1 \cdot \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 3 & 3 & 4 \\ \end{vmatrix} - (-1) \cdot \begin{vmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 3 & 3 & 3 \\ \end{vmatrix} \] But only the first term is nonzero due to zeros in the first row except the last element. #### The smaller determinant: \[ \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 3 & 3 & 4 \\ \end{vmatrix} \] Expand along the first row: \[ = 1 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 4 \\ \end{vmatrix} - 0 \cdot \begin{vmatrix} 0 & -1 \\ 3 & 4 \\ \end{vmatrix} + (-1) \cdot \begin{vmatrix} 0 & 1 \\ 3 & 3 \\ \end{vmatrix} \] \[ = 1 \cdot (1 \cdot 4 - (-1) \cdot 3) - 0 + (-1)[0 \cdot 3 - 1 \cdot 3] \] \[ = 1 \cdot (4 + 3) + 3 \] \[ = 7 + 3 = 10 \] ### Step 3: Final Value Thus, the determinant of matrix \( A \) is: \[ \boxed{10} \]