If the rank of the matrix `[[-1,2,5],[2,-4,a-4],[1,-2,a+1]]` is `1` then the value of `a` is (A) `-1` (B) 2 (C) `-6` (D) 4

Options

A. `2, if a=-6`
B. `2, if a=1`
C. `1, if a=2`
D. `1, if a=-6`

Explanation

To find the value of \( a \) such that the rank of the matrix

\[ \begin{bmatrix} -1 & 2 & 5 \\ 2 & -4 & a-4 \\ 1 & -2 & a+1 \end{bmatrix} \]

is 1, we will perform row operations and analyze the conditions for the rank.

### Step 1: Write down the matrix
The given matrix is

\[ A = \begin{bmatrix} -1 & 2 & 5 \\ 2 & -4 & a-4 \\ 1 & -2 & a+1 \end{bmatrix} \]

### Step 2: Perform row operations
We will perform the following row operations:
- Replace \( R_2 \) with \( R_2 + 2R_1 \)
- Replace \( R_3 \) with \( R_3 + R_1 \)

After performing these operations, we have:

1. \( R_1 \) remains unchanged:
\[ R_1 = \begin{bmatrix} -1 & 2 & 5 \end{bmatrix} \]

2. For \( R_2 \):
\[ R_2 = \begin{bmatrix} 2 & -4 & a-4 \end{bmatrix} + 2 \times \begin{bmatrix} -1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 0 & 0 & a + 6 \end{bmatrix} \]

3. For \( R_3 \):
\[ R_3 = \begin{bmatrix} 1 & -2 & a+1 \end{bmatrix} + \begin{bmatrix} -1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 0 & 0 & a + 6 \end{bmatrix} \]

Thus, the transformed matrix is:

\[ \begin{bmatrix} -1 & 2 & 5 \\ 0 & 0 & a + 6 \\ 0 & 0 & a + 6 \end{bmatrix} \]

### Step 3: Analyze the rank condition
For the rank of the matrix to be 1, there should be only one non-zero row. This means that the second and third rows must be zero:

\[ a + 6 = 0 \]

### Step 4: Solve for \( a \)
Solving the equation:

\[ a + 6 = 0 \implies a = -6 \]

### Conclusion
The value of \( a \) such that the rank of the matrix is 1 is

\[ \boxed{-6} \]