Step-by-step Solution:

The correct answer is \(\begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}\). To find the value of \(f(A)\), we first need to evaluate the powers of the matrix \(A\). This will reveal a helpful pattern. Step 1: Calculate \(A^2\) The first step is to compute the square of the matrix \(A\). Given \(A=\begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}\): $$A^2 = A \times A = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}$$ $$A^2 = \begin{bmatrix} (0)(0)+(5)(0) & (0)(5)+(5)(0) \\ (0)(0)+(0)(0) & (0)(5)+(0)(0) \end{bmatrix}$$ $$A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$ So, \(A^2\) is the zero matrix. A matrix whose power equals the zero matrix is known as a nilpotent matrix. Step 2: Simplify the Expression for \(f(A)\) The fact that \(A^2\) is the zero matrix greatly simplifies our problem. Any higher power of \(A\) will also be the zero matrix. * \(A^3 = A^2 \times A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \times A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) * Similarly, \(A^4, A^5, \dots, A^{16}\) will all be the zero matrix. Now we can simplify the expression for \(f(A)\): $$f(A) = I + A + A^2 + A^3 + \dots + A^{16}$$ $$f(A) = I + A + 0 + 0 + \dots + 0$$ $$f(A) = I + A$$ Step 3: Calculate the Final Result The final step is to add the identity matrix \(I\) and the matrix \(A\). $$f(A) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}$$ $$f(A) = \begin{bmatrix} 1+0 & 0+5 \\ 0+0 & 1+0 \end{bmatrix}$$ $$f(A) = \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}$$