Unknown 3x3 matrix

Question

It is my first time posting a question on this board, so I am sorry for my newbie mistakes, but I couldn't find a solution by searching around.

So my question is:

if an unknown $3\times3 $ Matrix $A$ is given the row operations $ero_1$, $ero_2$, $ero_3$, and $ero_4$ in the following order, it can be converted into the identitymatrix.

$$ero_1: -7r_1 + r_2 \implies r_2$$ $$ero_2: r_1 <-> r_3$$ $$ero_3: 1/8r_3 \implies r_3$$ $$ero_4: 12r_3 + r_1 \implies r_1$$

Determine for each $i = 1, 2, 3, 4$ the elementary matrix $E_i$, which is equivalent to the rowoperation $ero_i$ and justify $E_4E_3E_2E_1A = I$ applies.

Determine for each i = 1, 2, 3, 4 the elementary matrix $E_i^-1$

Determine the productmatrix $E_1^-1 E_2^-1 E_3^-1 E_4^-1$ and justify that it is equivalent to the unknown Matrix A.

So I get matrix A = $\pmatrix{0&0&8\\0&1&56\\1&0&-12}$

Does that look correct to you?

Any input is appreciated. Thank you.

Answer 1

Hint: $$ E_4E_3E_2E_1A = I \implies\\ E_4^{-1}E_4E_3E_2E_1A = E_4^{-1}I \implies\\ E_3E_2E_1A = E_4^{-1} \implies\\ E_2E_1A = E_3^{-1}E_4^{-1} \implies\\ A = E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1} $$ Or if you prefer, think about it this way: each row-operation is a step that can be reversed. Starting from the identity matrix, work backwards.

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Edit: The identity matrix of any size is simply the matrix with ones on the diagonal and zeros everywhere else. In this case, $$ I = \pmatrix{1&0&0\\0&1&0\\0&0&1} $$

Answer 2

Here's a different aspect from the realm of algebraic number theory. Let me tweak your whole thing for a moment. It fills in a lot of your missing proof-writing. I will say you are correct though.

Suppose that I is a unit (element whose reciprical is in the set we are working within) and there exist elements E and A such that E*A = I. Prove that E and A are units.

Essentially this is the problem you face. You dont know that E and A are unique... yet. Sure you might be able to say all the A's satisfy the equation and do some trickery but you still need to show A exists and is unique.

We'll assume A exists. Otherwise, we're dealing with absurdity. We know that E * A is a unit. Therefore, E is a unit and A is a unit. Therefore, E is invertible. All this does though is clear up the issue of existence.

You dont know the inverse of E is unique.

That is the problem really, and honestly I don't think that E is in general in possession of a unique inverse. Matrices have them, but I'm sure theres a field or ring where they don't. So I'll stop while Im ahead.

Your conclusion is correct. I didnt check the arithmetic of he matrix multiplication. I figure you're competent enough to do that. The method appears right and is right.