I am stuck at a step in some calculations from my teacher:
we have that $x_1$ and $x_2$ are non-zero and that $u_1>0$.
$ \begin{bmatrix} u_1 & 4 \\ 4 & u_1 \\ \end{bmatrix}$ * $\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}$ = $0$ $$-$$It seems like a trival step but the teacher uses :
$det\begin{bmatrix} u_1 & 4 \\ 4 & u_1 \\ \end{bmatrix}$ = $0$ $$-$$ How comes that we use the determinant to solve for our $u_1$?
On
Here is a justification. I change notation a bit. Suppose that we have a pair of equations $$ ax + by = 0\\cx+dy =0,\\$$ where $x$ and $y$ are not both zero. If $x =0$, then we have $by = 0$ and $dy =0$, so that $b = d = 0$ since $y \neq 0$, and the matrix $\left(\begin{array}{clcr} a& b\\c&d \end{array}\right)$ has zero determinant. So suppose that $x \neq 0$.
Multiply first equation by $d$ and second equation by $b$ to get: $$ adx + bdy = 0\\bcx+bdy =0.\\$$ Now subtract the second equation from the first to get $(ad-bc)x + 0y = 0$. Since $x \neq 0$, we must have $ad-bc = 0$ and the matrix $\left(\begin{array}{clcr} a& b\\c&d \end{array}\right)$ still has zero determinant.
On the other hand, if $ad-bc = 0$, we can solve the system of equations by taking $ x = d, y =-c$ and also by taking $x = -b, y = a$, so we have found at least one non-zero solution to the system of equations, unless $a = b = c = d = 0$. But if $a = b = c = d = 0$, then any choice of $x$ and $y$ not both zero gives a non-zero solution.
In general, if a square homogenous linear system has a non-zero solution, then the matrix cannot be invertible and so has determinant equal to zero.