Matrices - homogeneous system Hello I’m taking the course 18.02 multivariate calculus of MIT. I don’t understand the thinking behind the answer of a problem. 
The question is (with Julia notation): 
For what c-value(s) will $$\left(\begin{matrix}2& 1\\ 0 &-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right) = c\left(\begin{matrix}x\\y\end{matrix}\right)$$ have a non-trivial solution? (Write it as a system of homogeneous equations)
The answer is:
(1):    $\left(\begin{matrix}(2-c)x+y\\(-1-c)y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$ has a non-trivial solution if 
(2): $\left|\begin{matrix}(2-c)&1\\0&(-1-c)\end{matrix}\right|=0$ i.e., if:  
(3):    $(2-c)(-1-c) = 0$ or $c=2, c=-1$
I don’t understand how we come from the question to (1) or (2). Which theorem is used?
Thank you very much in advance
 A: $A\mathbf x = c\mathbf x\\
A\mathbf x - c\mathbf x = 0\\
A\mathbf x - cI\mathbf x = 0\\
(A\mathbf  - cI)\mathbf x = 0$
Either $\mathbf x = \mathbf 0$ (the trivial solution)
or $A - cI$ is singular, and $\det (A - cI) = 0$
A: We  have (1), which is $$\left(\begin{matrix}(2-c)x+y\\(-1-c)y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$
This can be rewritten as
$$\left(\begin{matrix}(2-c)&1\\0&(-1-c)\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$
There are now two ways to understand why the determinant of this matrix must be zero:
(a) - Suppose it does not have zero determinant. Then the matrix is invertible, so you can multiply by its inverse on both sides of this equation. Then you would get $(x,y)=(0,0)$, which is not a non-trivial solution (which is what you are asked for). So our assumption that it has non-zero determinant must be false.
(b) - You are looking for a non-trivial vector multiplied by a matrix to give the zero vector. This vector must then be an eigenvector of this matrix, with eigenvalue $0$. The determinant of a matrix is the product of the eigenvalues of the matrix, so if we have a $0$ eigenvalue, then the determinant must also be zero.
Therefore we get (2) $$\left|\begin{matrix}(2-c)&1\\0&(-1-c)\end{matrix}\right|=0$$
A: The theorem behind is Rouché–Capelli theorem which states that

A system of linear equations with n variables has a solution if and only if the rank of its coefficient matrix $A$ is equal to the rank of its augmented matrix $[A|b]$.If there are solutions, they form an affine subspace of $\mathbb {R} ^{n}$ of dimension $n − rank(A)$.
In particular:

*

*if $n = rank(A)$, the solution is unique,


*otherwise there is an infinite number of solutions.

In this case we are looking for no trivial solutions of $(A-cI)x=0$, that is for solutions with $x\neq 0$, and that require that $$rank(A)<n \iff \det(A-cI)=0$$
