Why is it that when a determinant = 0 then the homogeneous equations represented by the matrix has a non trivial solution? As I was following a lecture the instructor seemed to assume this and when on solve for the equations where the right side was equal to 0 and proceed with the problem but I know if a determinant is non zero than an inverse matrix exists and visa versa but I cannot seem to relate this to this.  In any event I cannot see the intuition as to why when the determinant is non zero then there are or is a non trivial solution.  I assume in this case the equations are homogeneous.  Thank you
 A: Let $M=(C_1,\cdots,C_n)$ an $n\times n$ matrix whose columns are $C_1,\cdots, C_n.$ If the determinant of $M$ is zero then their columns are linearly dependent. That is, there exists $a_1,\cdots, a_n\in\mathbb{R}$ (not all zero) such that $$a_1C_1+\cdots +a_nC_n=0.$$ Thus we have that
$$M\begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix}=0.$$ So $\begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix}$ is a non-trivial solution of $Mx=0.$
A: Take the system:
$$ \begin{cases} a x_1 + b x_2 = 0 \\ c x_1 + d x_2 = 0 \end{cases}$$
Note that you can write this in a matrix form:
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$
Note that this is an equation of the form $Ax=b$, where $b=0$.
The set of all vectors $x$ that, when multiplied by $A$ result in $0$ are called the null space of $A$ (actually we should talk about the transformations, but, for pedagogical purposes, i will just talk about the matrices now).
We know that, for some matrix $A$ that, if $\operatorname{det} A = 0$, then the matrix is not invertible. If the matrix is not invertible, then at least one of its rows can be expressed as a linear combination of the others. If that is the case, when solving the system, you will come to the fact that there is a free variable, and therefore, there is infinite solutions.
Therefore, if you don't have a row of such kind, then there is no free variable. If there is no free variable, then we have only 1 solution: and that must be the trivial solution.
