Question

Question from lecture 2: overview of linear algebra. How does Ax[1,0,0/-1,1,0/0,-1,1][x1/x2/x3] become x = [1,0,0/1,1,0/1,1,1][b1,b2,b3]. He says he takes the inverse? I don't follow how he came to this solution, especially the matrix part.

Answer 1

RKilly, I'm seeing that lecture 2 is "Elimination with matrices"; is that the one you're looking at? Roughly what time into the lecture does your question relate to?

Josh

Answer 2

I'm not sure which part of the lecture you're talking about but if I guess with the info in your question here's what I can summarise, hope it helps:

if Ax = b, where A is your first matrix [1 0 0; -1 1 0; 0 -1 1], x is your [x1, x2, x3] and b = [b1 b2 b3]

then what I think he was saying by combining your info here is that:

Since whatever times its inverse equals the identity matrix I of the form In, in a 3x3 matrix's case, I3= [1 0 0; 0 1 0; 0 0 1]

then A^-1 * A * x = I3* x = x = A^-1 * b

so x is just A^-1 * b, if you can find A^-1, then you can solve for x by multiplying A^-1 with the b column. This of course is way longer than just doing rref(A) but it's a valid method using the inverse.

Now to solve for A^-1 all you have to do is put the original matrix A in augmented format with the identity matrix I3 like this:
[1 0 0 | 1 0 0 ]
[-1 1 0 | 0 1 0 ]
[0 -1 1 | 0 0 1 ]

and do row operations to the Left Hand Side until it becomes the identity matrix and what you have on the right hand side of the augmented matrix will be your A^-1. which is the second matrix you listed in your question. x = [1,0,0/1,1,0/1,1,1][b1,b2,b3]

The reason this method works is because AxA^-1 = I, so whatever elimination methods we used to reduce A into I, it must be A^-1 itself, and this set of elimination operations are displayed on the augmented identity matrix since whatever we did to the left hand side, happened to the right hand side too.