YAMT:Finding two matrices A and B non-invt such that AB is invertible??

[Syringer]

AHH, I thought this would be a simple problem, but I can't seem to find it. Basically AB has to be a square matrix obviously..and I tried looking for A and B that aren't square matrices (and hence aren't invt), but I can't find them such that AB is invertible. Anyone have any ideas on examples?

 

[ed21x]

a= [1 0 1
1 1 1]

b= [1 1
1 0
1 1]

 

[ed21x]

i actually havn't taken linear algebra since sophomore year college, so my cross multiply on matrices might be shady =\

 

[Chebago]

whoa, I took Linear Algebra last semester and it still took me a minute, but yeah, you are right.

 

[Shooters]

Are you sure that A and B need not be square? The problem seems a little trivial if they're not.

I mean even A = [1 0] (row) and B = [1; 0] (column) would work.

Edit:
Nevermind the above comment. It wouldn't be possible if A and B were square.

 

[Syringer]

Originally posted by: ed21x
a= [1 0 1
1 1 1]

b= [1 1
1 0
1 1]

a*b results in:

2 1 2
1 0 1
2 1 2

And since row 1 and 3 are the same..it's rref is not In, and not invertible..

 

[Syringer]

Originally posted by: Shooters
Are you sure that A and B need not be square? The problem seems a little trivial if they're not.

I mean even A = [1 0] (row) and B = [1; 0] (column) would work.

Edit:
Nevermind the above comment. It wouldn't be possible if A and B were square.

A*B results in:

1 0
0 0

Which isn't invt..

 

[Shooters]

Your matrix multiplication is incorrect which is probably why you couldn't get an answer before.

 

[Syringer]

In fact for any matrix AxB for A:

a
b

and B:

c d

will not be invertible..

 

[Syringer]

Originally posted by: Shooters
Your matrix multiplication is incorrect which is probably why you couldn't get an answer before.

How so? What is AxB then?

 

[Shooters]

You're doing summation (column element of (a) * row element of (b)).

The correct way to do it is summation (row element of (a) * column element of (b)).

So for ed21x's example you should get

ab =
[2 2
3 2]

and for my example you should get

ab = [1]

both of which are invertible.

 

[Syringer]

I realize that, and that for

A=
[a
b], and
B=
[c d]

A*B =

[ac ad
cb bd]

Hence for your matrix, A*B=

[1 0
0 0]

 

[Shooters]

Look at ed21x's example and my example again.

In his example a is a 2x3 matrix and b is a 3x2 matrix. Therefore the product ab should be a 2x2, not a 3x3 like you said.

In my example a is a 1x2 and b is a 2x1. Hence the product should be a 1x1, not a 2x2.

 

[Syringer]

Ooooomg I'm such an idiot =/

Thanks for the help!

 

[Shooters]

Originally posted by: Syringer
Ooooomg I'm such an idiot =/

Thanks for the help!

No problem, it happens to all of us at some point in time.

 

[ed21x]

Originally posted by: Shooters

Originally posted by: Syringer
Ooooomg I'm such an idiot =/

Thanks for the help!

No problem, it happens to all of us at some point in time.

i simply prefer firing up matlab =)