solving linear systems inverse matrix
In school you solved systems of equations mechanically, by elimination or by substitution. You lined up two equations and knocked out x and y one at a time, probably without a feel for what that actually did. A system of equations really asks this: applying transformation A to some input produced b, so what was the input? It is the job of running a transformation backward. The lesson "Matrices Are Transformations" said a matrix is a transformation, and "The Determinant Is Area Scaling" said that transformation can squash space; here the two stories meet. When a system can be solved and when it cannot was in that picture all along.
Start by picturing the equation Ax = b. A is a transformation, x is the unknown input, and b is the result. Apply A to the plane and every point moves somewhere; the one point that lands exactly on b is the x we are looking for. In the figure, the gold ring is b, the blue arrow is a candidate input, and the green arrow is what A does to that candidate. Drag the blue dot until the tip of the green arrow sits in the gold ring: the arrow turns gold and the answer appears below. Solving a system of equations means working backward to the input that arrives at b.
Plotting candidates every time is impractical, so there is a smarter way. If A has bent space, there should be a transformation that exactly undoes the bending, and that is the inverse matrix A⁻¹. Step through the figure with the arrow buttons: the unit square is distorted by A and then returns to its original shape under A⁻¹. The box below shows the matrix currently applied, and in the last step it is the identity matrix, which changes nothing. So the answer is x = A⁻¹b. Pass b once through the transformation that runs backward, and you are at the starting point. The formula for solving a system was a rewind button all along.
Is there always exactly one answer? There is if the transformation has not squashed space, that is, if the determinant from the previous lesson is not 0. When the transformation spreads the points out evenly, exactly one input lands on any given result b. In the figure, the gold arrow is the result and the blue arrow is the input that produces it; drag the gold dot around and a single blue arrow follows every time. The line below shows the input's coordinates along with the determinant. For a transformation that can be undone there is always exactly one answer; that is what "a nonzero determinant means a clean solution" really says.
Now look at a determinant of 0. As the previous lesson showed, the plane is squashed onto a single line, and something curious happens. If the result b is not on that line, no input arrives there, so there is no answer. If b is on the line, countless inputs have been crushed onto that one point, so there are infinitely many answers. In the figure, the thick gray line is the squashed plane. Drag the arrow's tip onto that line and the label below reads "infinitely many"; drag it off and it reads "no solution." Either there is no answer or there are too many; those are the only options. What the determinant called "no inverse" shows up in a system of equations as "no clean answer."
Finally, connect this to the picture from school. Graph the two equations of a system and you get two lines; the answer is the point where they meet. The picture of two lines meeting is the same as the picture of running a transformation backward. Shrink the determinant with the slider and the green and orange lines become more and more parallel; the gold point where they cross is the answer x. The moment the determinant reaches 0, the lines are parallel and never meet. Then press the button below to make the right-hand sides agree, and the two lines coincide and meet everywhere. The first case has no answer and the second has infinitely many. A squashed transformation was a pair of parallel lines all along; the two pictures are really one.