Gaussian Elimination Geometrically
- Paul Pearson
Given two lines L1: 2x + y = 6 L2: 3x + 4y = 12 we construct a third line L3: a (2x + y) + b (3x + 4y) = 6a + 12b and vary the parameters a and b.
You should notice that L3 always goes through the point of intersection. In fact, if (x,y) is any point on both L1 and L2, then (x,y) satisfies both the equations for L1 and L2. Thus, if (x,y) is on both L1 and L2, the equation for L3 is true for all values of a and b; this is because if (x,y) is on both L1 and L2, then the expression 2x+y has the value 6 and the expression 3x+4y has the value 12 (even though we don't know what the values of x and y actually are yet, we know that these expressions have these values since (x,y) satisfies both L1 and L2). When we vary the parameters a and b, we can find some combinations that will either cancel the x-variable term in L3 (e.g., when a = 3 and b = -2) or cancel the y-variable term in L3 (e.g., when a = -4 and b = 1). Notice that changing the parameters a and b causes the line L3 to rotate about the point of intersection. The observations made here for pairs of lines in 2D space also generalize to planes in 3D space (and so on, to higher dimensions) as long as the system is consistent (i.e., there is at least one point of intersection of all of the lines in 2D, all the planes in 3D, etc.).