Exploring Linear Systems and Intersections of Planes

Intersections of planes A linear system is a mathematical model that can be described by an equation in which thevariables are unknown but are assumed to depend on one or more independent variablesx. A linear system can be classified as either homogeneous or inhomogeneous.Inhomogeneous systems have additional constraints, such as algebraic equations andinequalities, which must be satisfied by the solution of the system. Homogeneous systemsdo not have any such restrictions. So let's say we have a 3 by 3 linear system. Just to take an example, it doesn't reallymatter what we have, but let's say it is x plus z equals 1, x plus y equals 2, x plus 2y plus3z equals 3. Well, what does this mean? How do we solve it? It means we want to find x, yand z which satisfies all these conditions. So let's take a look at the first equation first. The first equation says that our point should be on the plane whose equation is given. Thesecond equation determines that our point should also be on another plane, since the twoplanes intersect. So in fact, these two equations give us two planes, since we know thatthe intersections of two planes are lines. Now, well, what happens with the third equation? That's actually going to be a third plane.In order to solve for the first two equations, we need to be on this line. If we want to solvefor a third equation, we also need to be on another plane. In general, three planes intersect in a point. This can be visualized by drawing linesthrough the first two planes of intersection and then projecting them onto the third plane tofind where they meet. And that point is the solution to the linear system. So the line, let's say, formed by theintersection between the first two planes intersects the third plane in a point, which is goingto be the solution. That was a mathematical notation for the intersection between the firsttwo planes. So, one way to solve this problem is to find the solution by drawing pictures and trying toidentify where the solution is. But we tend not to use this method in practice if we have theequations available. The solution is provided-- so let us use matrix notation. Remember, we saw before that the solution to AX = B is given by X = A inverse B. To getfrom here to here, multiply on the left by A inverse. A inverse × AX simplifies to X = Ainverse B. And once again, it's A inverse B and not BA inverse. If you try to set up themultiplication, BA inverse does not work. The sizes are not compatible. You cannotmultiply across or down.