Find all 2x2 matrices such that A=A^-1

1. Jul 5, 2012

autodidude

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

Find all 2x2 matrices such that $$A=A^-^1$$ (the inverse, just in case the notation is different)

2. Relevant equations

$$A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

3. The attempt at a solution

This is my second attempt at this question. The first time, I took a different approach, using the four equations below. This time, I started off with four other different equations and had a very tough time trying to figure it all out. If I hadn't looked at the solutions, I never would've gotten anywhere. Even with it, I struggled a lot and it took an unreasonable amount of time to work it all out...

----

If $$A=A^-1$$, then $$A^2=I$$. From that, I derived the following equations:

(i)$$a^2+bc=1$$
(ii)$$ab+bd=0$$
(iii)$$ac+cd=0$$
(iv)$$d^2+bc=1$$

From (ii) and (iii), if a≠-d, then b=0 and c=0. Then from (i) and (iv),a^2=1 and d^2=1and since a≠-d, then a=1 and d=1 or a=-1 and d=-1

So from that, I get two 2x2 matrices

\begin{bmatrix}
1 & 0
\\ 0&1
\end{bmatrix}

\begin{bmatrix}
-1 & 0
\\ 0&-1
\end{bmatrix}
[/tex]

Then to get the others, I let b=0 and c be any real number (again, I never would've done this if I hadn't seen the answers), then to satisfy (iii), a and d would have to be opposite. I did that for c=0 as well and the result was four more matrices

$$\begin{bmatrix} 1 & 0 \\ k &-1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ k &1 \end{bmatrix} \begin{bmatrix} 1 & k \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & k \\ 0&1 \end{bmatrix}$$

To get the last answer given in the book:

$$\begin{bmatrix} a & b \\ \frac{1-a^2}b&-a \end{bmatrix}$$

I just rearranged (i) to get b. But couldn't you write represent c in a similar way but with b in the denominator? Are they just writing it in terms of a and b?

Anyway, though it's solved, it took me super long and I'm still not happy with the method I used. I went down another road where I said b=0 iff a+d≠0 because I used

$$b=\frac{-b}{ad-bc}$$

$$ab+bd=0$$
$$b(a+d)=0$$
$$b=0$$

And I also said that if b≠0, then ad-bc-1 and I had all these equations but couldn't get anywhere with them.

Basically, I just want to see how you guys would approach this, what would be a systematic way to solve it? This wasn't even a challenge question and I got destroyed by it.

EDIT: Also, I'm unsure about my reasoning. I was more happy with the way I did it last time.

2. Jul 5, 2012

SammyS

Staff Emeritus
Subtract equation (iv) from equation (i).

That gives $d=\pm\,a\ .$

What do equations (ii) & (iii) tell you ?

3. Jul 5, 2012

InfinityZero

Well there is a quick way of finding the inverse of a 2x2 matrix

$$A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

$$A^{-1}=\frac{1}{det(A)}*adj(A)$$

and the adjugate of a 2x2 matrix is just

$$\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

the determinant is $$ad-bc$$

which by equating elements of the matrix and it's inverse you can show $$ad-bc=-1$$ and that $$d=-a$$

and from here getting the rest of the solution isn't too complicated.

4. Jul 6, 2012

Ray Vickson

Something you will find useful over and over again is a simple rule for inverting a 2x2 matrix:
swap the diagonal elements, change sign of the off-diagonal elements, and divide by the determinant. (Others have displayed this for you, but expressing it in words may help.)

RGV

5. Jul 6, 2012

I like Serena

Hi autodidude!

If you're interested in alternative and simpler ways to solve it, here's another (although it does require some advanced knowledge ).

Since you have $A^2-I=0$, according to Cayley–Hamilton, the eigenvalues of A must be roots of $x^2-1=0$.
So the eigenvalues can only be $\pm 1$.

And according to Jordan, the possible matrices A must then be "similar" to one of:
$$J = \begin{bmatrix}1&0\\0&1\end{bmatrix},\ \begin{bmatrix}-1&0\\0&-1\end{bmatrix},\ \begin{bmatrix}1&0\\0&-1\end{bmatrix},\ \begin{bmatrix}1&1\\0&1\end{bmatrix},\ \begin{bmatrix}-1&1\\0&-1\end{bmatrix}$$
However not all of these matrices are a solution: the last 2 do not "fit", which you can verify by calculating $A^2$, or by finding the inverse as mentioned before (swapping the entries on the main diagonal, negating the other 2 entries, and dividing by the determinant).

The term "similar" means that for any invertible matrix B, the matrix product $A = B J B^{-1}$ is a solution.

In the cases that J=I or J=-I, you can see that the corresponding matrices simplify to $A=B I B^{-1}=I$, respectively $A=B \cdot -I \cdot B^{-1}=-I$.

So the solutions with any invertible matrix B are:
$$A = I,\ A = -I,\ A = B \begin{bmatrix}1&0\\0&-1\end{bmatrix} B^{-1}$$

Last edited: Jul 6, 2012
6. Jul 12, 2012

autodidude

I did manage to get ad-bc = -1 the second time I did it. I think the part that wasn't supposed to be complicated turned out to be very complicated for me :)

@I like Serena: I'm very interested in other methods...but yeah, that's beyond where I'm at the moment. Looks much nicer :p