Start of with let g of f be a bijection, than state of the definition of a bijection. From there you can prove what must be true of g and f for g of f to meet the definition.
(A^-1) A^2 = (A^-1) ((a+d)A-(ad-bc)I)
(A^-1)(A)(A)= " " ""
IA = " " ""
A= (a+d)I-(ad-bc)A^-1
edit
A^-1 = 1/(ad-bc)(a+d)I- 1/(ad-bc)A
just check with calculator and it works.
Thanks Ray and Ivy.
Thanks for the reply.
We did that in class, I have it in my notes. I think the question is asking for something along this lines of
A^-1 = (b+a)A-(bc+da)I.
That isn't right as I just made it up, but that's the type of equation I think I suppose to come up with from this A^2 =...
1. a) Prove the following holds for A
A is a matrix [a b, c d]
I is identity matrix.
A^2 = (a+d)A-(ad-bc)I.
b) Assuming ad-bc not equal to 0, use a) to obtain an expression for A^-1.
The Attempt at a Solution
I proved the first equation, but I'm not seeing where it relates to...