Consider a system with 3 equations and 2 unknowns that has no solutions. List all possible arrangements of the 3 equations as lines on the x-y plane.
I know for a system to have no solution the determinant must be 0. But I do not understand what possible arrangements I can have.
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$\begingroup$Given that equations with two variables (let's assume $x,y \in \mathbb{R}$) are graphically lines on the plane $XY$ then the arrangements are the following:
- If there's one solution: the three lines intersect at the same point, which is the solution of the system. (The system is compatible and determined)
- There's no solution: in this case the three lines are somehow don't intersect at the same point. So any time they don't have a point in common it will be an incompatible system.
- They are all the same line: in that case they intersect in an infinite amount of points so there are infinite solutions to your system. (The system is compatible and undetermined).
An example of three (linear) equations in two unknowns is \begin{eqnarray*} x+2y&=&3 \\ 4x-5y&=&6 \\ -7x + 8y&=&9 \end{eqnarray*} Each of these equations gives a line in the plane. Three equations gives three lines in the plane.
How can three lines arrange themselves in the plane?
You could have three different parallel lines. You could have two different parallel lines and a transversal. You could have two concurrent lines and a third line parallel. You could have three non-parallel lines that all meet at a single point. There are many different possibilities.
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