Methods to finding multipicative inverses of matrices

[silkworm]

First of all, I'd like to apologize for how softball this question may be to this particular part of the forum, but I'm in a linear algebra course and I'm having difficulty with this particular part.

 

There is a matrix where A (A^-1) = I. I understand that, but I'm having difficulty finding a method for finding (A^-1). I thought I found one, but I worry that my book has a typo that throws the entire thing off. Any help on this would be greatly appreciated, and I'd like to mention I'm not just trying to leech, I've looked and can find nothing on methods, just the beginning and the end.

 

Thank you.

[Bignose]

See http://mathworld.wolfram.com/MatrixInverse.html for a pretty long discussion. There are easy formulas for 2x2 and 3x3 matrices, after that it gets complicated.

 

See especially the links in the above URL on Gauss-Jordan elimination, Gaussian elimination, and LU decomposition.

[matt grime]

Given a matrix A there is not always in inverse matrix. Sorry if you didn't mean that, but that's the implication I get from your post. I think googling for the adjugate matrix will help you.

[silkworm]

Thanks Bignose. That's a huge help. Oddly, I had to actually prove that particular thing in one of my problems (which I really couldn't do), but I had never been introduced to it before then.

 

matt grim, I know. It has to be nonsingular in order to have an inverse.

 

Thanks.

[silkworm]

Me again.

 

I'm working with a 3X3 now. The other was 2X2. I can see where the 1/(|A|) comes from with the 2X2. I can't see where it comes from with the 3X3. It looks as though the shift (or rearrangement of the matrices in the 2x2 that you multiply by the 1/|A|) is determined by the new matrix given by that formula, but I'm lost as to where the new |A| comes from (for the 3X3). I'm sure it's simple, but our teacher hasn't gone over it and I'm a spaz that has to take a test on this on Wednesday morning. I could figure it out, but I'm having trouble seeing what this number is. I mean, I understand it's purpose but it's origin is enigmatic to me at this point. I don't even see how the reciprocal can be determined by 1/ad-bc times the shift.

 

The first is like a cross product (for 2X2), right? Meaning it's orthagonal to both systems. And I can see that, but when it comes to the shift, that throws my thinking off because I don't understand the origin of the shift. So for the 3X3 I have no clue. Is it 1/(cross product of the 3X3) times the shift determined by the formula? And for the 3X3 they'd want (a3b2-b3a2, a1b3-b1a3, b2a1-a1b2) and then you find the square root of the sum of the squares of all 3 terms (Pythagorean?).

 

I'm sorry. I'm sure this is grade school for you guys, I'm just new.

[Connor]

an easy way to find A^(-1) (that I use) is to augment A with I, and put A:I in rref. you end up with I:A^(-1) (if A is independent, of course)

[silkworm]

an easy way to find A^(-1) (that I use) is to augment A with I, and put A:I in rref. you end up with I:A^(-1) (if A is independent, of course)

 

So you're saying set each row equal to the identity of that row? I'll have to figure out what independent and putting A:I in ref means. But is that what you're saying?

[Connor]

I'm sorry, I'm not too good at explaining things, but:

 

say you want to find the inverse of A

 

A=

|a b c|

|d e f |

|g h i |

 

augment I to A

 

A:I=

|a b c:1 0 0|

|d e f :0 1 0|

|g h i :0 0 1|

 

 

row reduce this matrix to:

 

|1 0 0:j k l |

|0 1 0:m n o |

|0 0 1:p q r |

 

the inverse of A is:

 

|j k l |

|m n o |

|p q r |

 

hope that helps

 

PS: rref means "reduced row echelon form"

[silkworm]

Thanks Connor. That makes perfect sense. I had to investigate it a little bit, but I totally dig it. I just have to get better at solving. The book keeping involved with these matrices are what is killing me now.