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Show that matrices of defined form have inverse of the same same defined form

 
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Sep22-10, 09:27 AM   #1
 

Show that matrices of defined form have inverse of the same same defined form


1. The problem statement, all variables and given/known data

Given the set of 3x3 matrices of the form: [1, a, b; 0, 1, c; 0, 0, 1], where a, b, and c are any real numbers show that the inverses of these matrices are of the same given form.

2. Relevant equations

Using elementary row operations, transform [A:I] into [I:A-1].
Inverse of a 3x3 matrix

3. The attempt at a solution

This is a subsection of a problem in which I am attempting to show that the set of these 3x3 matrices are a group under matrix multiplication. I was able to prove that it is well-defined, closed, an identity exists, and that associativity holds. For the inverse, it was simple to show that this set of 3x3 matrices is non-singular, but the trouble I'm running into is showing that the inverse is of the same given form so that closure still holds.

Thanks for any assistance.
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Sep22-10, 09:37 AM   #2
 
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Quote by donald17 View Post
1. The problem statement, all variables and given/known data

Given the set of 3x3 matrices of the form: [1, a, b; 0, 1, c; 0, 0, 1], where a, b, and c are any real numbers show that the inverses of these matrices are of the same given form.

2. Relevant equations

Using elementary row operations, transform [A:I] into [I:A-1].
Inverse of a 3x3 matrix

3. The attempt at a solution

This is a subsection of a problem in which I am attempting to show that the set of these 3x3 matrices are a group under matrix multiplication. I was able to prove that it is well-defined, closed, an identity exists, and that associativity holds. For the inverse, it was simple to show that this set of 3x3 matrices is non-singular, but the trouble I'm running into is showing that the inverse is of the same given form so that closure still holds.
What trouble are you having? Finding the inverse is straightforward, and yields the inverse in just a few steps. The inverse has the same form.
Sep22-10, 10:27 AM   #3
 
Actually I just solved it. Thanks.
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