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

autodidude

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

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

2. Relevant equations

[tex]

A=
\begin{bmatrix}
a & b
\\ c & d
\end{bmatrix}
[/tex]

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 [tex]A=A^-1[/tex], then [tex]A^2=I[/tex]. From that, I derived the following equations:

(i)[tex]a^2+bc=1[/tex]
(ii)[tex]ab+bd=0[/tex]
(iii)[tex]ac+cd=0[/tex]
(iv)[tex]d^2+bc=1[/tex]

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

[tex]
\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}
[/tex]

To get the last answer given in the book:

[tex]

\begin{bmatrix}
a & b
\\ \frac{1-a^2}b&-a
\end{bmatrix}
[/tex]

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

[tex]b=\frac{-b}{ad-bc}[/tex]

[tex]ab+bd=0[/tex]
[tex]b(a+d)=0[/tex]
[tex]b=0[/tex]

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.

SammyS

Subtract equation (iv) from equation (i).

That gives [itex]d=\pm\,a\ .[/itex]

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

InfinityZero

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

[tex]A=
\begin{bmatrix}
a & b
\\ c & d
\end{bmatrix}[/tex]

[tex]A^{-1}=\frac{1}{det(A)}*adj(A)[/tex]

and the adjugate of a 2x2 matrix is just

[tex]
\begin{bmatrix}
d & -b
\\ -c & a
\end{bmatrix}[/tex]

the determinant is [tex]ad-bc[/tex]

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

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

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

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}
$$

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