In this note, we discuss the relationship between a system of linear equations and the determinant of its coefficients. For simplicity, I am considering only a system of two linear equations with two variables. But a similar argument can be used for more general cases. Let us consider the system of linear equations $$\left\{\begin{aligned}ax+by&=e\\cx+dy&=f\end{aligned}\right.,$$ where none of $a,b,c,d,e,f$ is zero. The two linear equations are equations of lines in the plane, so we know there are three possibilities: There is no solution of the system in which case the two lines are parallel (so they do not meet), the system has a unique solution in which case the two lines meet at exactly one point, or the system has infinitely many solutions in which case the two lines are identical. This system can be written in terms of matrices as $$\begin{pmatrix}a & b\\c & d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}e\\f\end{pmatrix}$$ Let $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}$. If $\det A\ne 0$, then the system has a unique solution and it can be found using the Cramer’s rule as follows: $$x=\frac{\begin{vmatrix}e & b\\f & d\end{vmatrix}}{\det A},\ y=\frac{\begin{vmatrix}a & e\\c & f\end{vmatrix}}{\det A}$$ Note that $\det A=0$ if and only if the two lines have the same slope. Suppose that $\det A=0$. Then one can easily show that $\begin{vmatrix}e & b\\f & d\end{vmatrix}=0$ if and only if $\begin{vmatrix}a & e\\c & f\end{vmatrix}=0$. From $\det A=0$ and $\begin{vmatrix}e & b\\f & d\end{vmatrix}=0$, we have the system of equations: \begin{align}\label{eqn1}ad-bc&=0\\\label{eqn2}ed-fc&=0\end{align} Subtracting $a$ times \eqref{eqn2} from $e$ times \eqref{eqn1} yields $b(af-ec)=0$. Since $b\ne 0$, $af-ec=\begin{vmatrix}a & e\\c & f\end{vmatrix}=0$ which means that the two lines have the same $y$-intercept. This is the case when the two lines coincide and hence the system has infinitely many solutions (all the points on the line are solutions). Lastly, we know $\begin{vmatrix}e & b\\f & d\end{vmatrix}\ne0$ if and only if $\begin{vmatrix}a & e\\c & f\end{vmatrix}\ne0$. If $\begin{vmatrix}a & e\\c & f\end{vmatrix}\ne0$ while $\det A=0$, from the Cramer’s rule the system does not have a solution. $\begin{vmatrix}a & e\\c & f\end{vmatrix}\ne0$ means that the two lines have different $y$-intercepts, so this is the case when the two lines are parallel i.e. they do not meet. A system of homogeneous linear equations $$\left\{\begin{aligned}ax+by&=0\\cx+dy&=0\end{aligned}\right.$$ comes down to only two cases: the system has a unique solution $x=y=0$ (if $\det A\ne 0$) or has infinitely many solutions (if $\det A=0$). This is also obvious from considering two lines passing through the origin.