$A$ be a real symmetric non-singular matrix. Then which of the following is true?

Question

Let $A$ be a $n\times n$ real symmetric non-singular matrix. Suppose there exists $x\in \mathbb{R^n}$ such that $x'Ax<0$. Then we can conclude that

1. $\det(A)<0$
2. $B=-A$ is positive definite.
3. $\exists y\in \mathbb{R^n}: y'A^{-1}y<0$
4. $\forall y\in \mathbb{R^n}: y'A^{-1}y<0.$

My work:

I don't know how to do it for $n$, so I cooked up a $2\times 2$ matrix $\begin{bmatrix}-1 &2\\ 2& -1 \end{bmatrix}$, with $x=(1,0)^t$ we get $x'Ax<0$. Now $\det(A)=-3<0$ and also we can find a $y$ such that $y'A^{-1}y<0.$ But the problem is, according to the question only one option is true. And I don't know if this problem can be solved using a particular $n=2$.

So how can I solve this? Any help would be great. Thanks.

Answer 1

Take $n=2$,

For Option-$1$: Consider $A=\begin{bmatrix}-1 & 0\\ 0& -1 \end{bmatrix}$ then clearly $A$ is symmetric and non-singular matrix.

Moreover, there exists $x=(1,0)$ such that $x^TAx=-1$ but $det A=1$.

Option-$4$: Consider same matrix then $A^{-1}=\begin{bmatrix}-1 & 0\\ 0& -1 \end{bmatrix}$ but for $y=(0,0), y^TAy=0$

Option-$2$: Consider $A=\begin{bmatrix}-1 & 0\\ 0& 1 \end{bmatrix}$ then there exists $x=(-1,0)$ such that $x^TAx<0$ but $B=-A=\begin{bmatrix}1 & 0\\ 0& -1 \end{bmatrix}$ is not positive definite as determinant of $B$ is $-1$.

Hence, Option-$3$ is true.