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Solution of [A_{2x2}] = -

  1. Jun 4, 2014 #1

    BiGyElLoWhAt

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    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 :frown:
     
  2. jcsd
  3. Jun 4, 2014 #2

    Zondrina

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    [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
  4. Jun 4, 2014 #3

    BiGyElLoWhAt

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    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?
     
  5. Jun 4, 2014 #4

    LCKurtz

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    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.
     
  6. Jun 4, 2014 #5

    BiGyElLoWhAt

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    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.
     
  7. Jun 4, 2014 #6

    LCKurtz

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    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.
     
  8. Jun 4, 2014 #7

    BiGyElLoWhAt

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    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...
     
  9. Jun 4, 2014 #8

    LCKurtz

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    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?
     
  10. Jun 4, 2014 #9

    BiGyElLoWhAt

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    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 |##
     
  11. Jun 4, 2014 #10

    BiGyElLoWhAt

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    I see what you're (your XD) saying about his solution satisfying mine.
     
  12. Jun 4, 2014 #11

    LCKurtz

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    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.
     
  13. Jun 4, 2014 #12

    BiGyElLoWhAt

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    Thanks LCKurtz aka Guru of Matrices
     
  14. Jun 4, 2014 #13

    ehild

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    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
     
  15. Jun 5, 2014 #14

    LCKurtz

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