# Solution of [A_{2x2}] = -

1. Jun 4, 2014

### BiGyElLoWhAt

Solution of [A_{2x2}] = -

OK I'm not sure what my deal is with this.

My task is to find all solutions to $[A]_{2x2}^2 = -$

I 'know' what I'm looking for, and maybe that's part of my problem, as the teacher worked it out in class, but I didn't write down his solution because I didn't want to turn in his solution; I wanted to work it out for myself. Unfortunately, I think I'm overlooking something simple.

$\left | \begin{array}{cc} a & b \\ c & d \\ \end{array} \right | \times \left | \begin{array}{cc} a & b \\ c & d \\ \end{array} \right | = \left | \begin{array}{cc} -1 & 0 \\ 0 & -1 \\ \end{array}\right |$

which grants the equations

1: $a^2 + bc = -1$
2: $ab + bd = 0$
3: $ca + dc = 0$
4: $bc + d^2 = -1$

looking at eq.'s 2 & 3 we have:

$b(a+d) = 0 \ \text{&} \ c(a +d) = 0$

so we have 2 possible cases:
case 1:
$b=c=0$
if this is the case, substituting our values into eq. 1 and eq. 4 gives us
$a^2 = -1 \ \text{&} \ d^2 =-1$

which is what I did originally and talked to him about, but imaginary solutions are not allowed apparently. Inconsistent with the problem i.m.o. , but whatever.

so case 1 we give negative shits about.

case 2:
$a=-d$
and this is where I get stuck. It seems as though every substitution I make I get something to the effect of $0=0$ or something inconclusive, which I would need more linearly independent equations to solve.

$\left | \begin{array}{cc} 0 & -\frac{1}{c} \\ c & 0 \\ \end{array} \right |$

I guess my question is this:

Is it possible to work towards this solution without picking some arbitrary value for $a$ and thus $-d$? If there is, will someone point a finger in the direction I should be looking? I'm feeling really stupid right now

2. Jun 4, 2014

### Zondrina

[STRIKE]Alternatively, why not[/STRIKE] solve $[A]_{2 \times 2}^{2} + _{2 \times 2} = 0$;

For any arbitrary $A \in M_{2 \times 2} (\mathbb{R})$.

Last edited: Jun 4, 2014
3. Jun 4, 2014

### BiGyElLoWhAt

Hmmmm.... I'm not seeing how that's any different from my problem. I still need $[A]_{2x2}^2 = -$ to satisfy that equation.

But...

$\left | \begin{array}{cc} a & b \\ c & d \\ \end{array} \right | \times \left | \begin{array}{cc} a & b \\ c & d \\ \end{array} \right | + \left | \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right |$
grants the system:

1: $a^2 + bc + 1 = 0$
2: $ab + bd + 0 = 0$
3: $ca + cd + 0 = 0$
4: $bc + d^2 + 1 = 0$

...

Which is virtually the same system I had before.
Maybe I'm not getting the point.

Lets try multiplying both sides by $[A]^{-1}$

$[A]^{-1}[A][A] + [A]^{-1} = \vec{0}$
$[A] + [A]^{-1} = \vec{0}$
$[A] + [A]^{-1} = \vec{0}$

so...

$\left | \begin{array}{cc} a & b \\ c & d \\ \end{array} \right | + \frac{1}{ad-bc} \left | \begin{array}{cc} d & -c \\ -b & a \\ \end{array} \right | = \vec{0}$

which grants:

$a + \frac{d}{ad-bc} = 0$
$b - \frac{b}{ad-bc} = 0$
$c + \frac{c}{ad-bc} = 0$
$d - \frac{a}{ad-bc} = 0$

solving for a,b,c, and d:
$a = - \frac{d}{ad-bc}$
$b = \frac{b}{ad-bc}$
$c = -\frac{c}{ad-bc}$
$d = \frac{a}{ad-bc}$

make a substitution:

$a = - \frac{\frac{a}{ad-bc}}{ad-bc}$
which can only be satisfied if a = 0 or if $(\frac{1}{ad-bc})^2 = -1$ which isn't allowed, so a = 0, which I like. (hehe)

Which taking this solution and plugging it back into my original attempt gives me

$a^2 + bc = -1$
$(0)^2 + bc = -1$
$b=-\frac{1}{c}$

awesome

thanks for putting this in a new light for me. I literally have no idea why I didn't try multiplying by the inverse before, but whatever, I got it.

1 more question: this seems like a somewhat "round-about" way to solve this. Is there a simpler method that someone could point me towards?

4. Jun 4, 2014

### LCKurtz

Well, your prof's answer certainly isn't complete because$$A=\begin{bmatrix} 3 & 2\\ -5& -3 \end{bmatrix}$$certainly satisfies $A^2 = -I$. You are on the right track. You know you have to have $a=-d$ and when you do have that, equations 2 and 3 both work no matter what $c$ and $b$ are. What happens if you put $d=-a$ in the 4th equation? What does that tell you?

 When I said you are on the right track, I hadn't seen posts #2 and #3, which aren't.

5. Jun 4, 2014

### BiGyElLoWhAt

well when I plug $d= -a$ into equation 4 that gives me $bc + (-a)^2 = -1$ → $bc + a^2 = -1$ which is the exact same as equation 1. Which gives me a 0=0 that I mentioned earlier...

If somehow I could manipulate these to say that $-(a^2) = d^2$ that would make my life so much easier, but I'm not seeing how to do that either.

6. Jun 4, 2014

### LCKurtz

It doesn't give you $0=0$. It says the first and fourth equations are the same. So if you make the first equation work the fourth one automatically works. So at this point you have if $a = -d$ the second and third equations work for any $b$ and $c$. If you can figure out what $b$ and $c$ give you real values for $a$ in equation 1, everything will work. So think about figuring that out.

7. Jun 4, 2014

### BiGyElLoWhAt

OK, it only gives me 0=0 if i set equation 1 = to equation 4.

So what I'm gathering from you're posts LC, is that this can't be solved definitively. I'm gonna have something to the effect of b and c having opposite signs, as $-a^2-1 = bc$ and thus $bc$ must be negative. So c REALLY equals $-\frac{a^2+1}{b}$ and b REALLY equals $-\frac{a^2 +1}{c}$ so my general solution would thus be:
$\left | \begin{array}{cc} a & -\frac{a^2 +1}{c} \\ c & -a \\ \end{array} \right |$

Which satisfies the solution my prof had, but I though I was looking for something more specific. LOL fml

It's good to have some reassurance, but frustrating that I spent so much time looking for something that didn't exist because my prof was inconsistent with his own problem...

8. Jun 4, 2014

### LCKurtz

That's "your", the possessive, not a contraction for "you are". (Sorry, but that's a pet peeve of mine.)

Incorrect. It can be solved definitively. You just have to specify what values of the variables work.

Good.

Can $c=0$? (Why or why not?). I wouldn't say your solution satisfies your prof's solution. More the opposite, his is included in yours. But yours allows non-zeroes on the diagonal where he only has zeros, and the off diagonals don't have to be reciprocals. Also, as a matter of curiosity, does your solution include the example I gave?

9. Jun 4, 2014

### BiGyElLoWhAt

Well, no, c can't be 0, firstly that gives me a divide by zero, secondly it can't satisfy -a^2 - 1 = bc

and yes it does.

let a = 3 and c = -5:

$\left | \begin{array}{cc} 3 & (-\frac{9 + 1}{-5})=2 \\ -5 & -3 \\ \end{array} \right |$

10. Jun 4, 2014

### BiGyElLoWhAt

I see what you're (your XD) saying about his solution satisfying mine.

11. Jun 4, 2014

### LCKurtz

That's good. And I presume you see how to make zillions of other examples. That's the whole point of the exercise, to be able to describe all the matrices that work. Pick any $a$, pick any nonzero $c$ and presto, here's a matrix that works. And they all look like this.

12. Jun 4, 2014

### BiGyElLoWhAt

Thanks LCKurtz aka Guru of Matrices

13. Jun 4, 2014

### ehild

If the problem really was what you have written, your matrix is the complete solution if you add that a and c are arbitrary real numbers. It includes your Prof's solution (a=0) and also LCKurtz's one (with a=3 and c=-5).

ehild

ehild

14. Jun 5, 2014

### LCKurtz

Isn't that exactly what I said in post #11?