Matrix Algebra - fairly basic

• Oct 14th 2007, 09:43 AM
WWTL@WHL
Matrix Algebra - fairly basic
Using the definition of the inverse of a matrix and the principles of matrix algebra, prove that if A and B are invertible matrices of the same order, then AB is invertible and $\displaystyle (AB)^{-1}$ = $\displaystyle B^{-1} A^{-1}$.

• Oct 14th 2007, 10:06 AM
Jhevon
Quote:

Originally Posted by WWTL@WHL
Using the definition of the inverse of a matrix and the principles of matrix algebra, prove that if A and B are invertible matrices of the same order, then AB is invertible and $\displaystyle (AB)^{-1}$ = $\displaystyle B^{-1} A^{-1}$.

Hint: recall that if a matrix $\displaystyle A$ is invertible, and we denote $\displaystyle A^{-1}$ as its inverse matrix, then $\displaystyle AA^{-1} = I$ and $\displaystyle A^{-1}A = I$

here, we simply need to show that $\displaystyle (AB) \left( B^{-1}A^{-1}\right) = \left( B^{-1}A^{-1} \right)(AB) = I$, and the desired result follows immediately. use the fact that if $\displaystyle A,B,C,D$ are matrices, then $\displaystyle (AB)(CD) = A(BC)D$
• Oct 14th 2007, 10:15 AM
WWTL@WHL
http://www.mathhelpforum.com/math-he...f4c54e40-1.gif

Yes, that makes sense. But I don't understand why you'd need to use this fact http://www.mathhelpforum.com/math-he...485a3abe-1.gif. How does that relate to the http://www.mathhelpforum.com/math-he...f4c54e40-1.gif ?

Could you please expand on this a bit more, or give a more detailed explanation please? I am really, really poor at maths. :p

Thanks.
• Oct 14th 2007, 10:19 AM
Jhevon
Quote:

Originally Posted by WWTL@WHL
http://www.mathhelpforum.com/math-he...f4c54e40-1.gif

Yes, that makes sense. But I don't understand why you'd need to use this fact http://www.mathhelpforum.com/math-he...485a3abe-1.gif. How does that relate to the http://www.mathhelpforum.com/math-he...f4c54e40-1.gif ?

Could you please expand on this a bit more, or give a more detailed explanation please?

i give you one more hint.

Consider $\displaystyle (AB) \left( B^{-1}A^{-1} \right)$

in light of the fact that i told you to note (it would also do you well to recall the properties of the identity matrix), we will have that:

$\displaystyle (AB) \left( B^{-1}A^{-1} \right) = A \left(B B^{-1} \right)A^{-1}$

Quote:

I am really, really poor at maths. :p
you and me both
• Oct 14th 2007, 10:25 AM
WWTL@WHL
http://www.mathhelpforum.com/math-he...9ea2dfed-1.gif

$\displaystyle = A (I) A^{-1}$

$\displaystyle = (I) A A^{-1}$

$\displaystyle = I^2$ :confused:

It looks like I've committed a mathematical sin there, and I don't think I've got anywhere. :(

Just realised I^2 = I!! Thanks, I think I've got it.

Thanks so much Jhevon!!! :)
• Oct 14th 2007, 10:30 AM
Jhevon
Quote:

Originally Posted by WWTL@WHL
http://www.mathhelpforum.com/math-he...9ea2dfed-1.gif

$\displaystyle = A (I) A^{-1}$

$\displaystyle = (I) A A^{-1}$

$\displaystyle = I^2$ :confused:

It looks like I've committed a mathematical sin there, and I don't think I've got anywhere. :(

i told you to recall the properties of the identity matrix.

Definition: If we have a matrix, $\displaystyle I$, such that $\displaystyle IA = AI = A$ for any matrix $\displaystyle A$, then we call $\displaystyle I$ the identity matrix. It is a matrix of the same order of $\displaystyle A$ with a diagonal of $\displaystyle 1$'s on it's main diagonal and 0's everywhere else. Multiplying any matrix by the identity matrix gives the original matrix itself

in other words: $\displaystyle AIA^{-1} = AA^{-1}$
• Oct 14th 2007, 10:33 AM
WWTL@WHL
Quote:

Originally Posted by Jhevon
i told you to recall the properties of the identity matrix.

Definition: If we have a matrix, $\displaystyle I$, such that $\displaystyle IA = AI = A$ for any matrix $\displaystyle A$, then we call $\displaystyle I$ the identity matrix. It is a matrix of the same order of $\displaystyle A$ with a diagonal of $\displaystyle 1$'s on it's main diagonal and 0's everywhere else. Multiplying any matrix by the identity matrix gives the original matrix itself

in other words: $\displaystyle AIA^{-1} = AA^{-1}$

:o Yes - of course.

Thank you so much for your patience, and I'm sorry you had to walk me through this one. I really should've understood it. I'm only 2 weeks in to my maths course at university, and I'm already stuggling. Ho-hum.....:(
• Oct 14th 2007, 10:45 AM
Jhevon
Quote:

Originally Posted by WWTL@WHL
:o Yes - of course.

Thank you so much for your patience, and I'm sorry you had to walk me through this one. I really should've understood it. I'm only 2 weeks in to my maths course at university, and I'm already stuggling. Ho-hum.....:(

that's fine. i understand what it is to struggle with math, really. now can you complete the proof?
• Oct 14th 2007, 10:57 AM
WWTL@WHL
Quote:

Originally Posted by Jhevon
that's fine. i understand what it is to struggle with math, really. now can you complete the proof?

Yep. :)
• Oct 14th 2007, 11:00 AM
Jhevon
Quote:

Originally Posted by WWTL@WHL
Yep. :)

good!

(you can post it if you want to make sure everything is in order)