Expression for inverse matrix

1. Sep 15, 2011

Briane92

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.
3. The attempt at a solution
I proved the first equation, but I'm not seeing where it relates to the inverse. I know that ad-bc is the determinate. At first I was going to write A^-1 in terms of a,d,b,c in a matrix but I realize that this was done in class and its asking for an equation similar to the first one.

I just want a couple of hints, because I'm stuck.

2. Sep 15, 2011

HallsofIvy

Well, if
$$\begin{bmatrix}w & x \\ y & z\end{bmatrix}$$
is inverse to
$$\begin{bmatrix}a & b \\ c & d \end{bmatrix}$$
then we must have
$$\begin{bmatrix}w & x \\ y & z\end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}= \begin{bmatrix}aw+ cx & bw+ cd \\ ay+ cz & by+ cz \end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$

That gives you four equations to solve for w, x, y, and z.

3. Sep 15, 2011

Briane92

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 thats the type of equation I think I suppose to come up with from this A^2 = (a+d)A-(ad-bc)I equation. I did the work to show that is true. But I don't see the relation to the inverse except (ad-bc), the determinate, determines if A is invertible.

4. Sep 15, 2011

Ray Vickson

If B = A^(-1) exists, what do you get if you multiply your equation for A^2 by B on both sides?

RGV

5. Sep 15, 2011

Briane92

(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.

Last edited: Sep 15, 2011
6. Sep 15, 2011

Dick

Ok, so far. Now just solve that equation for A^(-1).

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