# Hessian matrix determinant greater than zero in a saddle point?

I have the function $$f:\mathbb{R}^3 \mapsto \mathbb{R}$$ defined as $$f(x, y, z) = xy+xz+yz+z-x$$. I've calculated the Jacobian:

$$J_f = (y+z-1, x+z, x+y+1)$$

Which, by setting $$J_f = \vec{0}$$, reveals that I have a stationary point at (-1, 0, 1). To find if it's a maxima or minima (or a saddle point) I derived the Hessian matrix. The Hessian matrix is a 3x3 symmetric matrix with zeroes in the main diagonal and ones in the other entries.

Now, $$det(H_f) = 0(0-1) - 1(0-1) + 1(1-0) = 2 > 0$$, hence this critical point should be a local maximum or minimum according to the second derivative test. However, the solution's manual sets this critical point as a saddle. Which one is correct?

The eigenvalues give me $$\lambda_1 = 2, \lambda_2 = \lambda_3 = -1$$.

• I think we must read the statement of the second derivative test more carefully. Apr 19 at 10:45

For functions of three or more variables, one needs to use the eigenvalues. In this case, as we have both positive and negative eigenvalues, $$(-1,0,1)$$ must be a saddle point.