Form signature from matrix

Determine the signature of the real quadratic form given by the matrix

Variant
\( \begin{pmatrix} 10 &1 \\ 1 &2 \end{pmatrix} \)
Solution
We perform Gaussian elimination by applying each transformation also on the corresponding column.

By subtracting the \(\frac1{10}\)-multiple of the first row from the second and then for the columns, we get:
\( \begin{pmatrix} 10 &1 \\ 1 &2 \end{pmatrix} \sim \begin{pmatrix} 10 &1 \\ 0 &\frac{19}{20} \end{pmatrix} \sim \begin{pmatrix} 10 &0 \\ 0 &\frac{19}{20} \end{pmatrix} \)
The signature is a triplet indicating the number of positive, negative and zero elements on the resulting diagonal matrix.
Alternative solution
The signature is also given by the signs of the eigenvalues of the matrix form. In our case, the characteristic polynomial is \(p_A(t)= \begin{vmatrix} 10-t &1 \\ 1 &2-t \end{vmatrix} = t^2-12t+19 \) and it has both roots positive.
Answer
The form signature is \((\#1, \#-1, \#0)=(2, 0, 0)\).
Variant
\(\begin{pmatrix} 1 &1 &-2 \\ 1 &0 &0 \\ -2 &0 &3 \\ \end{pmatrix}\)
Answer
The form signature is \((\#1, \#-1, \#0)=(2, 1, 0)\).
Variant
\( \begin{pmatrix} 1 &1 &-2\\ 1 &5 &0\\ -2 &0 &-4 \end{pmatrix} \)
Answer
The form signature is \((\#1, \#-1, \#0)=(2, 1, 0)\).
Variant
\( \begin{pmatrix} 1 &-1& 1\\ -1 &-3 &-3\\ 1 &-3 &0 \end{pmatrix} \)
Answer
The form signature is \((\#1, \#-1, \#0)=(1, 1, 1)\).