Question

4x^2+3y^2+z^2-8xy-6yz+4zx

Answer 1

Solution.

Write the given form as $4x^2-4xy+2xz-4xy+3y^2-3yz+2zx-3yz+z^2$

The corresponding matrix is $A=\begin{pmatrix} 4 & -4&2 \\ -4 & 3&-3\\ 2&-3&1 \end{pmatrix}$

We write $A=IAI$

We apply elementary operations on A and we apply the same row operations on the prefactor (R) and column operations on the post factor (C).

$\begin{pmatrix} 4 & -4&2 \\ -4 & 3&-3\\ 2&-3&1 \end{pmatrix}=\begin{pmatrix} 1 & 0&0 \\ 0&1&1\\ 0&0&1 \end{pmatrix}A\begin{pmatrix} 1 & 0&0 \\ 0&1&1\\ 0&0&1 \end{pmatrix}$

Applying $R_2+R_1, 2R_3-R_1, 2R_2-R_3,R_1-2R_3,R_1+R_2,R_1/2, R_2/\sqrt{2}, R_3/\sqrt{2},R_2\leftrightarrow R_3$

and $C_2+C_1, 2C_3-C_1, 2C_2-C_3, C_1-2C_3, C_1+C_2,C_1/2, C_2/\sqrt2,C_3/\sqrt2,C_2\leftrightarrow C_3$

we get

$\begin{pmatrix} 1 & 0&1 \\ 0 & -1&0\\ 0&0&-1 \end{pmatrix}=\begin{pmatrix} 3& 1&-3 \\ -1/\sqrt{2}&0&2/\sqrt2\\ 3/\sqrt2&2/\sqrt2&-2/\sqrt2 \end{pmatrix}A\begin{pmatrix} 3& -1/\sqrt2&3/\sqrt2 \\ 1&0&2/\sqrt2\\ -3&2/\sqrt2&-2/\sqrt2 \end{pmatrix}$

We have $D=P^TAP,$ where $D=\begin{pmatrix} 1 & 0&1 \\ 0 & -1&0\\ 0&0&-1 \end{pmatrix}.$

$P=\begin{pmatrix} 3& -1/\sqrt2&3/\sqrt2 \\ 1&0&2/\sqrt2\\ -3&2/\sqrt2&-2/\sqrt2 \end{pmatrix}.$

The linear transformation is X=PY, or

$\begin{pmatrix} x \\ y\\z \end{pmatrix}=\begin{pmatrix} 3& -1/\sqrt2&3/\sqrt2 \\ 1&0&2/\sqrt2\\ -3&2/\sqrt2&-2/\sqrt2 \end{pmatrix}\begin{pmatrix} y_1\\ y_2\\y_3 \end{pmatrix}$

From here

$$x=3y_1-1/\sqrt2 y_2+3/\sqrt2y_3,\newline y=y_1+\sqrt2 y_3,\newline z=-3y_1+\sqrt2 y_2-\sqrt2 y_3.$$

The canonical form is reduced to $X^TAX=Y^TDY$ .

We will have

$\begin{pmatrix} y_1&y_2&y_3 \\ \end{pmatrix}\begin{pmatrix} 1 & 0&1 \\ 0 & -1&0\\ 0&0&-1 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2\\y_3 \end{pmatrix}=y_1^2-y_2^2-y_3^2.$

Answer. The given quadratic form is reduced to normal form $y_1^2-y_2^2-y_3^2.$