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

[autodidude]

Homework Statement

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

Homework Equations

<br /> <br /> A=<br /> \begin{bmatrix}<br /> a &amp; b<br /> \\ c &amp; d<br /> \end{bmatrix}<br />

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

<br /> \begin{bmatrix}<br /> 1 &amp; 0<br /> \\ k &amp;-1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> -1 &amp; 0<br /> \\ k &amp;1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> 1 &amp; k<br /> \\ 0 &amp; -1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> -1 &amp; k<br /> \\ 0&amp;1<br /> \end{bmatrix}<br />

To get the last answer given in the book:

<br /> <br /> \begin{bmatrix}<br /> a &amp; b<br /> \\ \frac{1-a^2}b&amp;-a<br /> \end{bmatrix}<br />

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.

 

[SammyS]

Homework Statement

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

Homework Equations

<br /> <br /> A=<br /> \begin{bmatrix}<br /> a &amp; b<br /> \\ c &amp; d<br /> \end{bmatrix}<br />

The Attempt at a Solution

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
...

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

Subtract equation (iv) from equation (i).

That gives d=\pm\,a\ .

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

 

[InfinityZero]

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

A= <br /> \begin{bmatrix} <br /> a &amp; b <br /> \\ c &amp; d <br /> \end{bmatrix}

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

and the adjugate of a 2x2 matrix is just

<br /> \begin{bmatrix} <br /> d &amp; -b <br /> \\ -c &amp; a <br /> \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.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

<br /> <br /> A=<br /> \begin{bmatrix}<br /> a &amp; b<br /> \\ c &amp; d<br /> \end{bmatrix}<br />

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

<br /> \begin{bmatrix}<br /> 1 &amp; 0<br /> \\ k &amp;-1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> -1 &amp; 0<br /> \\ k &amp;1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> 1 &amp; k<br /> \\ 0 &amp; -1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> -1 &amp; k<br /> \\ 0&amp;1<br /> \end{bmatrix}<br />

To get the last answer given in the book:

<br /> <br /> \begin{bmatrix}<br /> a &amp; b<br /> \\ \frac{1-a^2}b&amp;-a<br /> \end{bmatrix}<br />

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.

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