$det(A)=0$ implies necessarily that if $X^T\cdot A\cdot X = 0$ so $A\cdot X = 0$?

Question

Let be $A$ a real $N\times N$ matrix and $X$ a real $N\times 1$ matrix.

1. If $X^T\cdot A \cdot X =0$ and $det(A)=0$, this implies necessarially that $A\cdot X = 0$?

2. If $X^T\cdot A \cdot X =0$ and $det(A)\neq 0$, exists a condiction to that $A\cdot X = 0$ be true ?

Answer 1

In both examples below, neither $X^TA$ nor $AX$ is zero: \begin{aligned} &\pmatrix{1&1&0}\pmatrix{1&0&0\\ 0&-1&0\\ 0&0&0}\pmatrix{1\\ 1\\ 0}=\pmatrix{1&1&0}\pmatrix{1\\ -1\\ 0}=0,\\ &\pmatrix{1&1}\pmatrix{1&0\\ 0&-1}\pmatrix{1\\ 1}=\pmatrix{1&1}\pmatrix{1\\ -1}=0. \end{aligned} One obvious sufficient condition that can force $AX$ to be zero whenever $X^TAX=0$ is that $A+A^T$ is positive definite, but this has little to do with $\det(A)$.