Thus, since $A$ is positive-definite, the matrix does not have $0$ as an eigenvalue.
Hence $A$ is invertible.

(b) Prove that $A^{-1}$ is symmetric.

By part (a), we know that $A$ is invertible. We have
\[A^{-1}A=I,\]
where $I$ is the $n\times n$ identity matrix.

Taking the transpose, we have
\begin{align*}
&I=I^{\trans}=(A^{-1}A)^{\trans}\\
&=A^{\trans}(A^{-1})^{\trans}\\
&=A(A^{-1})^{\trans} && \text{since $A$ is symmetric}.
\end{align*}
It follows that $A^{-1}=(A^{-1})^{\trans}$, and hence $A^{-1}$ is a symmetric matrix.

All eigenvalues of $A^{-1}$ are of the form $1/\lambda$, where $\lambda$ is an eigenvalue of $A$.
Since $A$ is positive-definite, each eigenvalue $\lambda$ is positive, hence $1/\lambda$ is positive.

So all eigenvalues of $A^{-1}$ are positive, and it yields that $A^{-1}$ is positive-definite.

Comment.

Note that the above proof of (b) shows the following.

The inverse matrix of a nonsingular symmetric matrix is symmetric.

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A real symmetric $n \times n$ matrix $A$ is called positive definite if \[\mathbf{x}^{\trans}A\mathbf{x}>0\] for all nonzero vectors $\mathbf{x}$ in...

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