1. **State the problem:** We are given a 5x5 matrix $$ \begin{bmatrix} 91 & 92 & 93 & 94 & 95 \\ 92 & 93 & 94 & 95 & 96 \\ 93 & 94 & 95 & 96 & 97 \\ 94 & 95 & 96 & 97 & 98 \\ 95 & 96 & 97 & 98 & 99 \end{bmatrix} $$ and we need to determine its rank. 2. **Recall the definition of rank:** The rank of a matrix is the maximum number of linearly independent rows or columns. 3. **Observe the matrix:** Each row is a sequence of consecutive integers increasing by 1, and each subsequent row is the previous row plus 1 element-wise. 4. **Check linear dependence:** Notice that the second row minus the first row is $$ \begin{bmatrix}92-91 & 93-92 & 94-93 & 95-94 & 96-95\end{bmatrix} = \begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix} $$ Similarly, the third row minus the second row is also $$\begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix}$$ This means all rows differ by the same vector, so the rows are linearly dependent. 5. **Express rows in terms of first two rows:** - Row 2 = Row 1 + $\begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix}$ - Row 3 = Row 2 + $\begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix}$ = Row 1 + 2$\begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix}$ - Similarly for rows 4 and 5. 6. **Check if the vector $\begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix}$ is linearly independent from Row 1:** Row 1 is $\begin{bmatrix}91 & 92 & 93 & 94 & 95\end{bmatrix}$, which is not a scalar multiple of $\begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix}$. 7. **Check if there is a third vector independent of these two:** Try to find if Row 3 can be written as a linear combination of Row 1 and Row 2. Since Row 3 = Row 1 + 2$\begin{bmatrix}1 & 1 & 1 & 1 & 1\end{bmatrix}$, it is dependent. 8. **Conclusion:** The rank is 2 because there are only two linearly independent rows. **Final answer:** $$\boxed{2}$$