Negative definite matrix and its inverse

Question

$$ A= \begin{pmatrix} -62 & -158 \\ -158 & -398\end{pmatrix} $$

Is $A$ negative definite?(Is ${\displaystyle z^{\textsf {T}}Az} <0 $ for every non-zero column vector $z$ of $n$ real numbers?)

What can we say above the definiteness of the inverse of $A$?

$$ A^{-1} = \frac{1}{144} \begin{pmatrix} -194 & -79 \\ -79 & 31 \end{pmatrix} $$

Is it negative definite as well?

Answer 1

Hint

Have you heard about eigenvalues of a square matrix? I suggest you to take a look at eigenvalues of a square matrix.

The sign of the eigenvalues of a matrix, always determines whether it is PSD (Positive Semi-Definite), PD, ND, NSD, ID.

Answer 2

$A$ is indefinite, since $\det A =-288 <0.$