The determinant `|(1,-2,3),(-4,5,-6),(-7,8,-9)|`

Answer

To find the value of the determinant \( |(1, -2, 3), (-4, 5, -6), (-7, 8, -9)| \), we will use the method of cofactor expansion.

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ D = \begin{vmatrix} 1 & -2 & 3 \\ -4 & 5 & -6 \\ -7 & 8 & -9 \end{vmatrix} \]

2. **Choose the First Row for Expansion**:
We will expand the determinant along the first row:
\[ D = 1 \cdot \begin{vmatrix} 5 & -6 \\ 8 & -9 \end{vmatrix} - (-2) \cdot \begin{vmatrix} -4 & -6 \\ -7 & -9 \end{vmatrix} + 3 \cdot \begin{vmatrix} -4 & 5 \\ -7 & 8 \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
- For the first determinant:
\[ \begin{vmatrix} 5 & -6 \\ 8 & -9 \end{vmatrix} = (5 \cdot -9) - (-6 \cdot 8) = -45 + 48 = 3 \]

- For the second determinant:
\[ \begin{vmatrix} -4 & -6 \\ -7 & -9 \end{vmatrix} = (-4 \cdot -9) - (-6 \cdot -7) = 36 - 42 = -6 \]

- For the third determinant:
\[ \begin{vmatrix} -4 & 5 \\ -7 & 8 \end{vmatrix} = (-4 \cdot 8) - (5 \cdot -7) = -32 + 35 = 3 \]

4. **Substitute Back into the Determinant**:
Now substitute the values of the 2x2 determinants back into the expression for \( D \):
\[ D = 1 \cdot 3 - (-2) \cdot (-6) + 3 \cdot 3 \]
\[ D = 3 - 12 + 9 \]

5. **Calculate the Final Value**:
\[ D = 3 - 12 + 9 = 0 \]

Thus, the value of the determinant is \( \boxed{0} \).