The rank of the matrix $$\begin{bmatrix}-1 & 2 & 5 \\2 & -4 & a - 4 \\1 & -2 & a + 1\end{bmatrix}\text{ is:} \\[12pt]$$​​

Let the given matrix be  A: 
​$$A = \begin{bmatrix}-1 & 2 & 5 \\2 & -4 & a - 4 \\1 & -2 & a + 1\end{bmatrix} \\[10pt]$$​​
We use Elementary Row Operations to find the Echelon Form.
​$$\text{Apply } R_2 \rightarrow R_2 + 2R_1 \text{ and } R_3 \rightarrow R_3 + 1R_1: \\$$​​
​$$\Rightarrow A \sim\begin{bmatrix}-1 & 2 & 5 \\0 & 0 & a + 6 \\0 & 0 & a + 6\end{bmatrix} \\[10pt]$$​​
Now apply $$R_3 \rightarrow R_3 - R_2: \\$$​​
​$$\Rightarrow A \sim\begin{bmatrix}-1 & 2 & 5 \\0 & 0 & a + 6 \\0 & 0 & 0\end{bmatrix} \\[10pt]$$​​
The rank of  A is the number of non-zero rows in this echelon form.
Rank is 2 if there are exactly two non-zero rows. This requires: 
​$$a + 6 \neq 0 \Rightarrow a \neq -6 \\[6pt]$$​​
Thus, the rank is $$2 if a \neq -6$$​​
$$\therefore$$​The rank is 2 if  a = 1.