The value of the determinant `|{:(1,0,-1),(a,1,1-a),(b,a,1+a-b):}|` depends on :

Options

A. only a
B. only b
C. neither a nor b
D. both a and b

Answer

To solve the determinant \( D = \begin{vmatrix} 1 & 0 & -1 \\ a & 1 & 1-a \\ b & a & 1+a-b \end{vmatrix} \), we will expand it using the first row.

### Step 1: Expand the determinant using the first row
We can expand the determinant along the first row. The formula for the determinant expansion is:

\[ D = a_{11} \cdot D_{11} - a_{12} \cdot D_{12} + a_{13} \cdot D_{13} \]

where \( D_{ij} \) is the determinant of the submatrix obtained by deleting the \( i \)-th row and \( j \)-th column.

In our case:
- \( a_{11} = 1 \), \( a_{12} = 0 \), \( a_{13} = -1 \)

Thus, we have:

\[ D = 1 \cdot D_{11} + 0 \cdot D_{12} - 1 \cdot D_{13} \]

### Step 2: Calculate \( D_{11} \) and \( D_{13} \)

1. **Calculate \( D_{11} \)**:
\[ D_{11} = \begin{vmatrix} 1 & 1-a \\ a & 1+a-b \end{vmatrix} = (1)(1+a-b) - (1-a)(a) = 1 + a - b - a + a^2 = a^2 + 1 - b \]

2. **Calculate \( D_{13} \)**:
\[ D_{13} = \begin{vmatrix} a & 1 \\ b & a \end{vmatrix} = (a)(a) - (1)(b) = a^2 - b \]

### Step 3: Substitute back into the determinant formula

Now substituting back into our expression for \( D \):

\[ D = 1 \cdot (a^2 + 1 - b) - 1 \cdot (a^2 - b) \]

### Step 4: Simplify the expression

Now simplify:

\[ D = (a^2 + 1 - b) - (a^2 - b) = a^2 + 1 - b - a^2 + b \]

Notice that \( -b \) and \( +b \) cancel out, and \( a^2 \) and \( -a^2 \) also cancel out:

\[ D = 1 \]

### Conclusion

The value of the determinant \( D \) is always \( 1 \), regardless of the values of \( a \) and \( b \). Therefore, the determinant depends on neither \( a \) nor \( b \).