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Math Help - Logical ways to find certain matrices...

  1. #1
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    Logical ways to find certain matrices...

    I'm stuck on 2 problems. It's not the answers I care about, just the understanding of how to get to the answer.

    1) Can you find a 3x2 matrix A and a 2x3 matrix B such that AB = I?

    2) Find a 2x2 matrix A =/= I with A^3 = I.

    How can you do these without brute forcing the solution?
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  2. #2
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    Quote Originally Posted by paulrb View Post
    I'm stuck on 2 problems. It's not the answers I care about, just the understanding of how to get to the answer.

    1) Can you find a 3x2 matrix A and a 2x3 matrix B such that AB = I?

    2) Find a 2x2 matrix A =/= I with A^3 = I.

    How can you do these without brute forcing the solution?
    1) For a matrix B to be the inverse of matrix A, it must satisfy the following equation:

     AB = I = BA

    So basically, if you can make up a random non-singular matrix and call it A, then B will be its inverse, and I'm sure you know how to find that!

    2)  A^3 = A^2 \times A = I

    So now you have to find a matrix A whose inverse is also its square!

     A = \left( \begin{array}{ccc}<br />
a & b  \\<br />
c & d \end{array} \right)

     A^2 =  \left( \begin{array}{ccc}<br />
a & b  \\<br />
c & d \end{array} \right) \left( \begin{array}{ccc}<br />
a & b  \\<br />
c & d \end{array} \right)=  \left( \begin{array}{ccc}<br />
a^2+b^2 & ab+bd  \\<br />
ac+cd & d^2+bc \end{array} \right)

     A^{-1} = \frac{1}{ad-bc} \left( \begin{array}{ccc}<br />
d & -b  \\<br />
-c & a \end{array} \right)

    Hence:

     \frac{d}{ad-bc} = a^2+b^2 , etc. 4 equations, 4 unknowns, you do the math.
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  3. #3
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    For #1 I'm pretty sure there's no such thing as a non-square matrix that has an inverse (such that AB = I = BA).

    For #2, I know the square of the matrix is also the inverse of the matrix. But it still seems kind of unintuitive how to arrive at the solution...
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  4. #4
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    Quote Originally Posted by paulrb View Post
    For #1 I'm pretty sure there's no such thing as a non-square matrix that has an inverse (such that AB = I = BA).

    For #2, I know the square of the matrix is also the inverse of the matrix. But it still seems kind of unintuitive how to arrive at the solution...
    Ah sorry, I didn't quite read your first question as carefully as I should have!
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  5. #5
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    Ah ok well the second one makes sense now. Thanks
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  6. #6
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    Quote Originally Posted by paulrb View Post
    Ah ok well the second one makes sense now. Thanks
    Alternatively just find  A^3 and equate it to the identity, as opposed to finding the square and the inverse and equating them...

    For the first one, set up two matrices with random variables again:

     A = \left( \begin{array}{ccc}<br />
a & b & c \\<br />
d & e & f \end{array} \right)


     B = \left( \begin{array}{ccc}<br />
g & h  \\<br />
i & j  \\<br />
k & l  \end{array} \right)

     AB = \left( \begin{array}{ccc}<br />
ag+bi+ck & ah+bj+cl  \\<br />
dg+ei+fk & dh+ej+fl  \end{array} \right)

    We know this must equate to:

     AB = \left( \begin{array}{ccc}<br />
1 & 0  \\<br />
0 & 1  \end{array} \right)

    So, if we say that b = i = c = k = 0, and a = g = 1, then the first element is satisfied. Now look at the send element:

    ah + bj + cl = 1h + 0j + 0l = 0, hence h = 0.

    Now look at the 3rd element: dg+ei+fk = d(1) + e(0) + f(0) = 0, hence d= 0.

    Now the last element: pmath]dh+ej+fl = 0(0) + e(0) + f(l) = 1[/tex], so just set f = l = 1, and you are sorted.

    Also e = j = 0

    Hence:

     A = \left( \begin{array}{ccc}<br />
1 & 0 & 0 \\<br />
0 & 0 & 1 \end{array} \right)


     B = \left( \begin{array}{ccc}<br />
1 & 0  \\<br />
0 & 0  \\<br />
0 & 1  \end{array} \right)
    Last edited by Mush; April 2nd 2009 at 06:29 PM.
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