# 2x2 matrix inverse formula

1. Sep 24, 2008

### hoffmann

I need to find the inverse of a 2x2 matrix [a b ; c d] using Gauss-Jordan elimination.

I am halfway there but I'm stuck on the algebra because it gets really messy. Could anyone possibly do it step by step?

2. Sep 24, 2008

### gabbagabbahey

Why don't you show us what you've got so far?

3. Sep 24, 2008

### hoffmann

sure:

[ a b ; c d | 1 0 ; 0 1 ] -->
[ a b ; (ac/c) (ad/c) | 1 0 ; 0 (a/c) ] -->
[ a b ; 0 ((ad/c)/c) -b | -1 (a/c) ] -->
...

here's where i'm a little stuck. i'm bad at keeping track of every variable...i think i miss something along the way because of the messy algebra.

4. Sep 24, 2008

### gabbagabbahey

Assuming that your last line is supposed to be:
$$\begin{pmatrix} a & b &1 & 0 \\ 0 & \frac{ad}{c}-b & -1 & \frac{a}{c} \end{pmatrix}$$

5. Sep 24, 2008

### hoffmann

here it is:

look good?

6. Sep 24, 2008

### gabbagabbahey

Are you multiplying the top row by (ad/c-b)/b ? If so, you should get:
$$\begin{pmatrix} \frac{a(\frac{ad}{c}-b)}{b} & (\frac{ad}{c}-b) &\frac{(\frac{ad}{c}-b)}{b} & 0 \\ 0 & \frac{ad}{c}-b & -1 & \frac{a}{c} \end{pmatrix}$$

7. Sep 24, 2008

### hoffmann

ah right, so the next step is:

it's messy this way...sorry.

8. Sep 24, 2008

### gabbagabbahey

Wouldn't the step be to subtract the bottom row from the top row to get:
$$\begin{pmatrix} \frac{a(\frac{ad}{c}-b)}{b} & 0 &\frac{(\frac{ad}{c}-b)}{b}+1 & \frac{-1}{c} \\ 0 & \frac{ad}{c}-b & -1 & \frac{a}{c} \end{pmatrix}=\begin{pmatrix} \frac{a(\frac{ad}{c}-b)}{b} & 0 &\frac{ad}{bc} & \frac{-1}{c} \\ 0 & \frac{ad}{c}-b & -1 & \frac{a}{c} \end{pmatrix}$$

9. Sep 24, 2008

### hoffmann

alright, so now we have a matrix with zeros along the anti-diagonal. the inverse doesn't equal the inverse given by the 2x2 inverse formula. what went wrong?

10. Sep 24, 2008

### gabbagabbahey

You still have to set the diagonal elements to 1: simply multiply the top row by b/(a(ad/c-b)) and the bottom row by 1/(ad/c-b)