Proving the Inverse of a 2x2 Matrix Using Row Operations

[forty]

Use row operations to show that the inverse of the 2*2 matrix:

[ a , b ]
[ c , d ]is the matrix (1/(ad-bc))[ d , -b ]
[ -c , a ]

provided that ad-bc =/= 0I created an augmented matrix as follows

[ a , b | 1 , 0 ]
[ c , d | 0 , 1 ]

so no i have to use row operations to make the left hand side an identity matrix...

divide row1 by a and row 2 by c

[ 1 , b/a | 1/a , 0 ]
[ 1 , d/c | 0 , 1/c ]

then row2 minus row1[ 1 , b/a | 1/a , 0 ]
[ 0 , d/c - b/a | -1/a , 1/c ]But as what to do from here I'm stumped everything i try just ends up leading me in a circle... like usual any help would be greatly appreciated :)

 

[Vanadium 50]

I first said "Since they give you the inverse, wouldn't it be easier to just multiply the matrix by its inverse and see what you get?" but I see they want you to do it with row operations.

Step one is get the lower left element zero. You've done that. Your next step will be to make the upper right element zero. That's your goal.

 

[Hurkyl]

[ 1 , b/a | 1/a , 0 ]
[ 0 , d/c - b/a | -1/a , 1/c ]

But as what to do from here I'm stumped everything i try just ends up leading me in a circle... like usual any help would be greatly appreciated :)

The basic row reduction algorithm is purely algorithmic; you simply carry out the individual steps and you get the right answer¹. I don't understand how you can be going in circles -- what is it you were 'trying'?



1. Assuming you haven't made a mistake

 

[forty]

You where right, i wasn't actually going in circles just had to keep following the algebra through and it all fell out in the end.

(a friend did it this way: http://img255.imageshack.us/img255/466/scan0003jy7.jpg which was a lot simpler then the way i did it by starting with r1/a and r2/c)

Thanks a lot :)

 

[Hurkyl]

You where right, i wasn't actually going in circles just had to keep following the algebra through and it all fell out in the end.

Ah. That makes sense. I know what it's like to lose confidence in the middle of a calculation -- but it's frequently worth continuing just to make sure! (Especially if it's a calculation that's supposed to work!)

One general trick is to find alternate ways to judge your progress. In this particular case, when doing row reduction, you can always measure progress by how many columns you've reduced. The point at which you stopped you had fully reduced one column, so you were half-way done. (actually, less than halfway, because the later steps take less work)