The quadratic form can be written as

$$\begin{bmatrix} x_{1} & x_{2} & x_{3} \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$$

$\therefore A = \begin{bmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{bmatrix}$

We know $A = IAI \\ \begin{bmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ R_{3} \rightarrow R_{3} - \frac{1}{3} R_{1}, C_{3} \rightarrow C_{3} - \frac{1}{3} C_{1} \\ \begin{bmatrix} 3 & -1 & 0 \\ -1 & 5 & -\frac{2}{3} \\ 0 & - \frac{2}{3} & \frac{8}{3} \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{1}{3} & 0 & 1 \end{bmatrix} A \begin{bmatrix} 1 & 0 & -\frac{1}{3} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ R_{2} \rightarrow 3R_{2} + R_{1}, C_{2} \rightarrow 3C_{2} + C_{1} \\ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 14 & -2 \\ 0 & -2 & \frac{8}{3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 3 & 0 \\ -\frac{1}{3} & 0 & 1 \end{bmatrix} A \begin{bmatrix} 1 & 1 & -\frac{1}{3} \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ R_{3} \rightarrow 7R_{3} + R_{2}, C_{3} \rightarrow 7C_{3} + C_{2} \\ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 14 & -2 \\ 0 & 0 & \frac{50}{3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 3 & 0 \\ -\frac{4}{3} & 3 & 7 \end{bmatrix} A \begin{bmatrix} 1 & 1 & -\frac{4}{3} \\ 0 & 3 & 3 \\ 0 & 0 & 7 \end{bmatrix} \\ R_{1} \rightarrow \frac{1}{\sqrt{3}}R_{1}, C_{1} \rightarrow \frac{1}{\sqrt{3}}C_{1} \\ R_{2} \rightarrow \frac{1}{\sqrt{14}}R_{2}, C_{2} \rightarrow \frac{1}{\sqrt{14}}C_{2} \\ R_{3} \rightarrow \sqrt{\frac{3}{50}}R_{3}, C_{3} \rightarrow \sqrt{\frac{3}{50}}C_{3} \\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{sqrt{3}} & 0 & 0 \\ \frac{1}{\sqrt{14}} & \frac{3}{\sqrt{14}} & 0 \\ -\frac{4}{\sqrt{150}} & 3 \sqrt{\frac{3}{50}} & 7 \sqrt{\frac{3}{50}} \end{bmatrix} A \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{14}} & - \frac{4}{\sqrt{150}} \\ 0 & \frac{3}{\sqrt{14}} & 3 \sqrt{\frac{3}{50}} \\ 0 & 0 & 7\sqrt{\frac{3}{50}} \end{bmatrix}$

Thus $B = P'AP$ where B is diagonal matrix.

$X'AX = 3x_{1}^2 + 5x_{2}^2 + 3x_{3}^2 - 2x_{1}X_{2} - 2x_{2} x_{3} + 2x_{3}x_{1}$ will be transformed to

$Y'By = u^2 + v^2 + w^2$

By transformation X = PY

$x = \frac{1}{\sqrt{3}}u + \frac{1}{\sqrt{14}}v - \frac{4}{\sqrt{150}}w, \hspace{1cm} y = \frac{3}{\sqrt{14}}v + 3\sqrt{\frac{3}{50}}w, \hspace{1cm} z = 7\sqrt{\frac{3}{50}}w$

Rank(r): 3

Signature(s) = difference between positive squares and negative squares = 3 - 0 = 3

Index: No. of positive squares = 3(Also r = s)