Find the determinant of the \( n \times n \) matrix \( A \) with 5 's on the diagonal, 1 's above the diagonal, and 0 's below the diagonal. (Note that your answer will be a function of \( n \).) \( \operatorname{det}(A)= \)

To find the determinant of the matrix \( A \), we can denote it as an upper triangular matrix with 5's on the diagonal and 1's above the diagonal. The matrix takes the form: \[ A = \begin{pmatrix} 5 & 1 & 1 & \cdots & 1 \\ 0 & 5 & 1 & \cdots & 1 \\ 0 & 0 & 5 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 5 \end{pmatrix} \] Using properties of determinants, specifically for triangular matrices, the determinant is the product of the diagonal entries. However, our configuration has additional non-zero entries above the diagonal. We can utilize a recursive approach or apply a pattern recognition with determinants of smaller sizes. Let’s denote \( d_n = \det(A) \) for an \( n \times n \) matrix \( A \). We can establish a relationship based on the size of the matrix. We consider expanding the determinant: 1. For \( n = 1 \), \( d_1 = 5 \). 2. For \( n = 2 \), the matrix becomes \( A = \begin{pmatrix} 5 & 1 \\ 0 & 5 \end{pmatrix} \). - Thus, \( d_2 = 5 \cdot 5 = 25 \). For larger matrices, we could use the Laplace expansion or other properties, but we note that the determinant relates recursively to \( d_{n-1} \). Generally, the pattern continues as: \[ d_n = 5 d_{n-1} - d_{n-2} \] With initial conditions \( d_1 = 5 \) and \( d_2 = 25 \). Solving this recurrence relationship gives a general expression. Ultimately, the solution to the recurrence yields: \[ \det(A) = 5^n - (n - 1) \cdot 1^{n} = 5^n - (n-1) \] Thus, we conclude: \[ \operatorname{det}(A) = 5^n - (n-1) \]