Can Gauss-Jordan Elimination Help Me Find the Inverse of a 2x2 Matrix?

[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?

 

[gabbagabbahey]

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

 

[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.

 

[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}

then your doing fine so far. what is your next step?

 

[hoffmann]

here it is:

[ (a(((ad/c)-b)/b) (((ad/c)-b)/b) ; 0 ((ad/c) -b) | (((ad/c)-b)/b) 0 ; -1 (a/c) ]

look good?

 

[gabbagabbahey]

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

 

[hoffmann]

ah right, so the next step is:

[ (a(((ad/c)-b)/b) - (((ad/c)-b)) 0 ; 0 ((ad/c) -b) | ((((ad/c)-b)/b) - (ad/c) -b) 0 ; -1 (a/c) ]

it's messy this way...sorry.

 

[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} &amp; 0 &amp;\frac{(\frac{ad}{c}-b)}{b}+1 &amp; \frac{-1}{c} \\ 0 &amp; \frac{ad}{c}-b &amp; -1 &amp; \frac{a}{c} \end{pmatrix}=\begin{pmatrix} \frac{a(\frac{ad}{c}-b)}{b} &amp; 0 &amp;\frac{ad}{bc} &amp; \frac{-1}{c} \\ 0 &amp; \frac{ad}{c}-b &amp; -1 &amp; \frac{a}{c} \end{pmatrix}

 

[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?

 

[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)