Proving ad-bc ≠ 0: RREF Matrix Solution | Step-by-Step Guide

[icesalmon]

Homework Statement

Show that if ad-bc ≠ 0 then
RREF[a b] = [1 0]
[c d] [0 1]

The Attempt at a Solution

I cannot use any equivalent statements here, and I don't "know" what a determinant is yet so I have to show this in another way.

I start off by saying if ad - bc ≠ 0 then I can say ad - bc = "e" whether it's positive or negative, I'm not sure, is important. So what I tried was using some substitution and got
a = ( e + bc ) / d
b = ( ad -e ) / c
c = ( ad - e ) / b
d = ( e + bc ) / a
and creating a little 2 x 2 matrix with the RHS, it got too messy and I wasn't even sure what I was doing would work so I stopped with that.

then I went back to ad - bc = ±e and tried working on some ideas with inequalities for instance if ad - bc = +e then I know ad > bc and if ad - bc = -e then I know ad < bc and this is where I get stuck. I'm not sure if any of this is on the right track, but as always, I've come to PF to untangle my mangle. Thanks for the help.

The little scribble at the end of the problem statement is supposed to be a 2x2 matrix row reducing to the identity.

 

[Mark44]

Homework Statement

Show that if ad-bc ≠ 0 then
RREF[a b] = [1 0]
[c d] [0 1]

The two matrices are
$$ \begin{bmatrix} a & b\\ c & d\end{bmatrix}$$
and
$$ \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$$

The Attempt at a Solution

I cannot use any equivalent statements here, and I don't "know" what a determinant is yet so I have to show this in another way.

I start off by saying if ad - bc ≠ 0 then I can say ad - bc = "e" whether it's positive or negative, I'm not sure, is important. So what I tried was using some substitution and got
a = ( e + bc ) / d
b = ( ad -e ) / c
c = ( ad - e ) / b
d = ( e + bc ) / a
and creating a little 2 x 2 matrix with the RHS, it got too messy and I wasn't even sure what I was doing would work so I stopped with that.

then I went back to ad - bc = ±e and tried working on some ideas with inequalities for instance if ad - bc = +e then I know ad > bc and if ad - bc = -e then I know ad < bc and this is where I get stuck. I'm not sure if any of this is on the right track, but as always, I've come to PF to untangle my mangle. Thanks for the help.

The little scribble at the end of the problem statement is supposed to be a 2x2 matrix row reducing to the identity.

You don't need to worry about ad - bc being positive or negative; just whether it ad - bc = 0 or ad - bc ≠ 0. What I would do is actually row-reduce your first matrix. You should see in a short time what effect the value of ad - bc has on what you get.

 

[icesalmon]

okay so I'm row reducing \begin{bmatrix} a & b\\ c & d\end{bmatrix}
and I get\begin{bmatrix} (ad-bc) & 0\\ 0 & (ad-bc)\end{bmatrix}
and I'm concerned about ad - bc and it's relationship with 0. Well if ad - bc = 0 then we don't get the 2x2 identity, so how do I show it's exactly equal to one and not some other number?
Am I just setting:
\begin{bmatrix} (ad-bc) & 0\\ 0 & (ad-bc)\end{bmatrix}
equal to
\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} and calling it finished?

 

[Mark44]

okay so I'm row reducing \begin{bmatrix} a & b\\ c & d\end{bmatrix}
and I get\begin{bmatrix} (ad-bc) & 0\\ 0 & (ad-bc)\end{bmatrix}
and I'm concerned about ad - bc and it's relationship with 0. Well if ad - bc = 0 then we don't get the 2x2 identity, so how do I show it's exactly equal to one and not some other number?

You don't have to show that ad - bc = 1. All you need to know is that ad - bc ≠ 0. As long as that's true, you can multiply each row by 1/(ad - bc), which gives you 1 for the upper left and lower right entries in the matrix.

Am I just setting:
\begin{bmatrix} (ad-bc) & 0\\ 0 & (ad-bc)\end{bmatrix}
equal to
\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} and calling it finished?

 

[Mark44]

There are three row operations that you can use to row-reduce a matrix:
1. Adding a multiple of one row to another.
2. Interchaning two rows.
3. Replacing a row by a nonzero multiple of itself.

The matrix that you end up with after doing these operations is equivalent to, but not equal to, the one you started with.

 

[icesalmon]

so based on the condition, that we assume from the start, which is ad - bc isn't equal to 0 we can divide out the ( ad - bc ) terms to make it equal to the identity without having to worry about any impossibilities: 1/0
working off of #3 there.

It would help if I were to pay more attention to the problem statement from now on.
Thanks, Mark

 

[Mark44]

so based on the condition, that we assume from the start, which is ad - bc isn't equal to 0 we can divide out the ( ad - bc ) terms to make it equal to the identity without having to worry about any impossibilities: 1/0

Pretty much.

working off of #3 there.

It would help if I were to pay more attention to the problem statement from now on.

That's always good advice!

Thanks, Mark

You're welcome