| written 9.4 years ago by |
The quadratic form can be written as
$$\begin{bmatrix} x_{1} & x_{2} & x_{3} \end{bmatrix} \begin{bmatrix} 1 & -2 & 5 \\ -2 & 1 & -1 \\ 5& -1 & -2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$$
$\therefore A = \begin{bmatrix} 1 & -2 & 5 \\ -2 & 1 & -1 \\ 5& -1 & -2 \end{bmatrix}$
We know $A = IAI$
$\begin{bmatrix} 1 & -2 & 5 \\ -2 & 1 & -1 \\ 5& -1 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ R_{3} \rightarrow R_{3} - 5 R_{1} , C_{3} \rightarrow C_{3} - 5C_{1}, R_{2} \rightarrow R_{2} + 2R_{1} , C_{2} \rightarrow C_{2} + 2C_{1} \\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & -3 & 9 \\ 0 & 9 & -27 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -5 & 0 & 1 \end{bmatrix} A \begin{bmatrix} 1 & 2 & -5 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ R_{3} \rightarrow R_{3} + 3R_{2}, C_{3} \rightarrow C_{3} + 3C_{2} \\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 3 & 1 \end{bmatrix} A \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix} \\ R_{2} \rightarrow \frac{1}{\sqrt{3}}R_{2}, C_{2} \rightarrow \frac{1}{\sqrt{3}} C_{2} \\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 7 0 & 0 \\ \frac{2}{\sqrt{3}} & \frac{1}{\sqrt{3}} & 0 \\ 1 & 3 & 1 \end{bmatrix} A \begin{bmatrix} 1 & \frac{2}{sqrt{3}} & 1 \\ 0 & \frac{1}{\sqrt{3}} & 1 \\ 0 & 0 & 1 \end{bmatrix} $
Thus $B = P'AP$ where B is diagonal matrix
$X'AX = x^2 + y^2 + 2z^2 - 4xy - 2yz + 10xz$ will be transformed to
$Y'BY = u^2 - v^2$
By transformation X = PY
$x = u + \frac{2}{\sqrt{3}}v + w \\ y = \frac{1}{\sqrt{3}}v + 3w \\ z = w$
Rank:2
Signature = difference between positive squares and negative squares = 2 - 1 = 1
Index: No. of positive squares = 1
Value class is indefinite

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