I have the following matrix
$$A = \begin{pmatrix} 2 & 5 & 8 \\ 3 & 6 & 9 \\ 4 & 7 & 9 \end{pmatrix}$$
I already calculated the determinant with Laplace in two different ways, row and column expansion. Now I was going to perform a Gauss reduction, and then use the scale factor to show the determinant is the same.
By the way, det$(A) = 3$.
Yet after having done Gauss algorithm, the scale factor doesn't add properly. I think I'm blind and I don't see something really obvious...
These are the operations:
- multiply row $2$ ( = $R_2$) by $\frac{1}{3}$
$$A = \begin{pmatrix} 2 & 5 & 8 \\ 1 & 2 & 3 \\ 4 & 7 & 9 \end{pmatrix}$$
- Computing $2R_2 - R_1$ in the place of $R_2$ and $R_3- 2R_1$ in the place of $R_3$:
$$A = \begin{pmatrix} 2 & 5 & 8 \ 0 & -1 & -2 \ 0 & -3 & -7 \end{pmatrix}
- Multiply by $(-1)$ the last two rows
$$\begin{pmatrix} 2 & 5 & 8 \\ 0 & 1 & 2 \\ 0 & 3 & 7 \end{pmatrix}$$
- Last: $R_3 - 3R_2$
$$A = \begin{pmatrix} 2 & 5 & 8 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}$$
So the scale factor is
$$d = \frac{1}{3}\cdot (-1) \cdot (-1) = \frac{1}{3}$$
yet $$\text{det}(A) = 2d \rightarrow \frac{2}{3}$$
What I am not seeing?
SOLVED