Look at the matrix $$A=\begin{pmatrix} -1& 0& -1\\ 0&-2&0\\ 1&0&4 \end{pmatrix}.$$ It is the Hessian matrix of a certain function in 3-variables evaluated at a point $P$. In order to understand if $P$ is a relative minimum or a maximum point, I have to study the matrix $A$ by using the pivots. In particular, if all the pivots are strictly positive, the Hessian matrix is positive definite and then the point $P$ is a minimum.
Now, in order to have the matrix in echelon form, I proceed with the elementary operation of substituting the 3rd row with the sum between the first and the third one, obtaining the matrix $$\begin{pmatrix} -1& 0& -1\\ 0&-2&0\\ 0&0&3 \end{pmatrix}.$$ Since we have 2 positive pivots and one is negative, thus the matrix is neither positive definite nor negative definite, thus $P$ is a saddle point.
My question is: In these kind of exercise, in order to obtain the echelon form of the matrix, only linear substitutions of row with their linear combinations are allowed (e.g. permutation of rows are not allowed!). Why I can't, for example, multiply the third row by $-1$ obtaining $$\begin{pmatrix} -1& 0& -1\\ 0&-2&0\\ 0&0&-3 \end{pmatrix}$$ and say it is negative definite?
Could anyone please help me?
Thank you in advance!
The determinant of a square matrix is unchanged by adding one row to another row, since
The solution consists of the row operation $R_3 \leftarrow R_3 + R_1$, so the determinant of the given matrix is unchanged by this row operation.
However, by multiplying a row of a square matrix by a constant $k$, the new determinant becomes $k$ times the determinant of the original square matrix, due to the multilinearity of the determinant.
In particular, when you multiply the third row by $-1$, you've changed the sign of the determinant, so that would give you the opposite of what you want.