Inverse Matrices?

Question

I am given a 2-by-2 matrix A (0,8 / 5,7) which I am asked to write as a product of elementary matrices and then write the inverse of A as aproduct of elementary matrices. So I turned A into reduced row echlon form using the following steps:

-Switch R1 and R2
-1/5 * R1
-1/8 * R2
- (-7/5) * R2 + R1 into R1

So the elementary matrices are:
(0,1 / 1,0) * (1/5,0 / 0,1) * (1,0 / 1/8,0) and (1,-7/5 / 0,-7/5)

Right?

Then the inverse of A as a product of elementary matrices is just the inverse of each of (1,-7/5 / 0,-7/5) * (1,0 / 1/8,0) * (1/5,0 / 0,1) and (0,1 / 1,0).

Does this make sense? Because I keep getting the wrong answer and I don't know where I've gone wrong....

Answer 1

Well, you've made several mistakes. One of them is that you're getting the elementary matrices wrong. Your transposition matrix is okay, as is the matrix for multiplying row 1 by 1/5, but the matrix for multiplying row 2 by 1/8 is [[1, 0], [0, 1/8]], not [[1, 0], [1/8, 0]]. Remember, the matrix for multiplying row n by k has k in the nth row and nth column (read: on the diagonal), and 1's elsewhere on the diagonal. Similarly, your row addition matrix is screwed up -- it should be [[1, -7/5], [0, 1]], not [[1, -7/5], [0, -7/5]]. Remember, when you add a multiple of row n to row m, row n must remain unchanged.

It may help to remember that the elementary matrix for a given row operation is precisely the matrix that would result if you applied that operation to the identity matrix.

With that out of the way, let me talk about your second mistake. You're computing the wrong elementary matrices.

Let's think of this algebraically. You have a matrix A, and you transform it into matrix B by performing an elementary row operation on it. Performing a row operation on A is equivalent to left-multiplying it by the corresponding elementary matrix E. So B is the result of left-multiplying A by E, so B = EA. Does it follow that A=EB? Of course not -- A = E⁻¹B. In other words, when factorizing a matrix into elementary matrices, you should record the matrix of the opposite transformation, not the one you just did. So the factorization of A would be:

[0, 1][5, 0][1, 0][1, 7/5]
[1, 0][0, 1][0, 8][0, 1]

You may easily verify that the product of those matrices is in fact A. The inverse of A, as a product of elementary matrices, would then be the product of the inverses of those matrices, in reverse order. So we have for A⁻¹:

[1, -7/5][1, 0][1/5, 0][0, 1]
[0, 1] [0, 1/8] [0, 1] .[1, 0]

Or multiplying these out:

[-7/40, 1/5]
[1/8, 0]

Which you may easily verify is indeed A⁻¹.

Answer 2

Let Ei - the ith elementary matrix...

identitiy = (En)(En-1) ... (E3)(E2)(E1)A

this is the supposed relationship...
just check again...

btw... the last elementary matrix should be
(1 , -7/5 / 0 , 1)

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