1. **Problem:** Determine the rank of the matrix $$\begin{bmatrix} 91 & 92 & 93 & 94 & 95 \\ 92 & 93 & 94 & 95 & 96 \\ 93 & 94 & 95 & 96 & 97 \\ 94 & 95 & 96 & 97 & 98 \\ 95 & 96 & 97 & 98 & 99 \end{bmatrix}$$ 2. **Formula and rules:** The rank of a matrix is the maximum number of linearly independent rows or columns. We use row operations to reduce the matrix to row echelon form and count the nonzero rows. 3. **Step 1:** Observe the matrix is a Toeplitz matrix with each row shifted by 1. 4. **Step 2:** Subtract row 1 from row 2: $$R_2 \to R_2 - R_1 = \begin{bmatrix} 92-91 & 93-92 & 94-93 & 95-94 & 96-95 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \end{bmatrix}$$ 5. **Step 3:** Similarly, subtract row 2 from row 3: $$R_3 \to R_3 - R_2 = \begin{bmatrix} 93-92 & 94-93 & 95-94 & 96-95 & 97-96 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \end{bmatrix}$$ 6. **Step 4:** Notice rows 2 and 3 are identical after subtraction, so they are linearly dependent. 7. **Step 5:** Subtract row 2 from row 3: $$R_3 \to R_3 - R_2 = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ 8. **Step 6:** Similarly, subtract row 3 from row 4 and row 4 from row 5, we get zero rows. 9. **Step 7:** So, the matrix has only 2 linearly independent rows. 10. **Answer:** The rank of the matrix is $2$.