Convert the quadratic form to canonical form and determine the transformation matrix.
$f(x_1, x_2, x_3)$ = $x_1^2 + x_2^2 + 2x_1x_2 - 2x_1x_3 + 6x_2x_3$
So I reduced the form to the following form: $(x_1 + x_2 - x_3)^2 - (2x_2 + x_3)^2 + 4x_2^2$ and substituted:
$y_1 = x_1 + x_2 - x_3$
$y_2 = -2x_2 - x_3$
$y_3 = 4x_2$
I thought that just successive rows of this transition matrix should be formed by the values of $y_1, y_2$ and $y_3$. And the matrix would look like this:
\begin{bmatrix}
1 & 1 & -1 \\
0 & -2 & -1 \\
0 & 4 & 0 \\
\end{bmatrix}
But someone told me that I should solve $x_1, x_2, x_3$ with $y_1, y_2, y_3$ and the correct matrix will be:
\begin{bmatrix}
1 & 1 & \frac{1}{4} \\
0 & \frac{1}{4} & 0 \\
0 & -1 & -\frac{1}{2} \\
\end{bmatrix}
What is correct?