The set of matrices that are their own inverse in R2

[Elwin.Martin]

Homework Statement

Find all 2x2 square matrices A which are their own inverses.

Homework Equations

A²=I
A=A^-1

The Attempt at a Solution

I know that the diagonal is comprised of 1s and or -1s and the other entries are zero but I can't seem to show it algebraically.

I went the weak way and did components...
I said A had rows of {a,b} and {c,d} and that A²=I
so
a²+bc=1
b(a+d)=0
c(a+d)=0
d²+bc=1

Now I know that the solution will have diagonals comprised of (+ or -)1 and the other two entries zero but I don't know how to do the algebra to get there =| I'll write down what I tried last night, any direction would be greatly appreciated!

So I said that
b(a+d)=0
c(a+d)=0
so either (a+d)=0 or b=0 and c=0...
so if we assume a+d=!0 then
b=0 and c=0
thus d²=1
and a²=1
so we'd have a=+-1,b=0,c=0,d=+-1

but if a+d=0
then a=-d
d²-a²=0
d=+-a but apparently the positive case is ignored since we assumed a+d=0 s0 d=-a again and I have nothing conclusive about b or c?

Again, thank you in advanced for any and all advice...it's sad that my basic algebra skills are so weak...if someone could recommend a good algebra practice book that would also be appreciated. I think I'm worse at Algebra than everything else haha

Also if this thread needs to be moved just let me know, at my university Linear Algebra is beyond Calculus but this particular question is fairly simple.

Thank you again for your time
elwin

 

[micromass]

Homework Statement

Find all 2x2 square matrices A which are their own inverses.

Homework Equations

A²=I
A=A^-1

The Attempt at a Solution

I know that the diagonal is comprised of 1s and or -1s and the other entries are zero but I can't seem to show it algebraically.

I went the weak way and did components...
I said A had rows of {a,b} and {c,d} and that A²=I
so
a²+bc=1
b(a+d)=0
c(a+d)=0
d²+bc=1

Now I know that the solution will have diagonals comprised of (+ or -)1 and the other two entries zero but I don't know how to do the algebra to get there =|

Why do you think these are all the matrices? This isn't true.

I'll write down what I tried last night, any direction would be greatly appreciated!

So I said that
b(a+d)=0
c(a+d)=0
so either (a+d)=0 or b=0 and c=0...
so if we assume a+d=!0 then
b=0 and c=0
thus d²=1
and a²=1
so we'd have a=+-1,b=0,c=0,d=+-1

but if a+d=0
then a=-d
d²-a²=0
d=+-a but apparently the positive case is ignored since we assumed a+d=0 s0 d=-a again and I have nothing conclusive about b or c?

So you know you have a=-d. And you also know that a²+bc=1. So you can express b in terms of a and c...

 

[Elwin.Martin]

Why do you think these are all the matrices? This isn't true.

Oh wow...that changes things quite a bit...

So you know you have a=-d. And you also know that a²+bc=1. So you can express b in terms of a and c...

Thank you very much! I think I know what I'm doing wrong now.
elwin