What Are All the Solutions for the Matrix Equation A² = -I for 2x2 Matrices?

[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. what the prof worked to in class was this:$\left | \begin{array}{cc}
0 & -\frac{1}{c} \\
c & 0 \\
\end{array} \right |$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

 

[STEMucator]

[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})##.

 

[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?

 

[LCKurtz]

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. what the prof worked to in class was this:$\left | \begin{array}{cc}
0 & -\frac{1}{c} \\
c & 0 \\
\end{array} \right |$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

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?

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

 

[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.

 

[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.

 

[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 going to 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...

 

[LCKurtz]

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,

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

is that this can't be solved definitively.

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

I'm going to have something to the effect of b and c having opposite signs, as $-a^2-1 = bc$ and thus $bc$ must be negative.

Good.

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

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?

 

[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:

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

 

[BiGyElLoWhAt]

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

 

[LCKurtz]

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:

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

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.

 

[BiGyElLoWhAt]

Thanks LCKurtz aka Guru of Matrices

 

[ehild]

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 going to 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...

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

 

[LCKurtz]

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

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