• Support PF! Buy your school textbooks, materials and every day products Here!

Invertible square matrix

  • Thread starter Precursor
  • Start date
  • #1
222
0

Homework Statement


Show that the square matrix [tex]2N - I[/tex] is its own inverse if [tex]N^{2} = N[/tex]

Homework Equations


properties of invertible matrix


The Attempt at a Solution


I really don't know where to start here. I know that [tex](2N-I)(2N-I) = I[/tex], but where do I go on from there?
 

Answers and Replies

  • #2
614
0
You are correct in calculating (2N-I)(2N-I) and using N2=N in your calculations. If (2N-I)(2N-I)=I, then the desired result is achieved, as this implies [(2N-I)]-1=(2N-I).
 
  • #3
222
0
You are correct in calculating (2N-I)(2N-I) and using N2=N in your calculations. If (2N-I)(2N-I)=I, then the desired result is achieved, as this implies [(2N-I)]-1=(2N-I).
So the answer is simply [tex](2N-I)(2N-I) = I[/tex]? But how did I use [tex]N^{2} = N[/tex]?
 
  • #4
614
0
I just assumed you did :)
The result (2N-I)(2N-I)=I is the desired result. So take (2N-I)(2N-I) and expand it. After you are done that, use N2=N.
 
  • #5
222
0
I just assumed you did :)
The result (2N-I)(2N-I)=I is the desired result. So take (2N-I)(2N-I) and expand it. After you are done that, use N2=N.
ok, so I get:

[tex]2N^{2} - 2NI - 2NI + I^{2} = I[/tex]

[tex]2N - 2N - 2N + I = I[/tex]

But this doesn't work out.
 
  • #6
614
0
The coefficient on N2 is incorrect. It is not a 2.
 
  • #7
222
0
The coefficient on N2 is incorrect. It is not a 2.
Ok I got it. Thanks for the help.
 
  • #8
614
0
Cheers.
 

Related Threads on Invertible square matrix

  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
997
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
895
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
3K
Top