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Math Help - Matrix problem, please help!

  1. #1
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    Exclamation Matrix problem, please help!

    I have the following problem:

    I have 2 3x3 matrices, on called M and the other N^-1 (this one has a few unknown numbers, k). I am asked to find the following:

    (MN)^-1. Now we have NOT learned how to do the inverse of 3x3 matrices yet, so there must be some other way of solving this...but how?

    M=
    (3 0 8
    0 2 0
    1 0 4)
    and N^-1=
    (1 -2 3
    3 k 0
    2 4 k)

    Please help...
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  2. #2
    MHF Contributor
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    West Malaysia
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    Quote Originally Posted by HelenaStage View Post
    I have the following problem:

    I have 2 3x3 matrices, on called M and the other N^-1 (this one has a few unknown numbers, k). I am asked to find the following:

    (MN)^-1. Now we have NOT learned how to do the inverse of 3x3 matrices yet, so there must be some other way of solving this...but how?

    M=
    (3 0 8
    0 2 0
    1 0 4)
    and N^-1=
    (1 -2 3
    3 k 0
    2 4 k)

    Please help...
    hi

     <br />
(MN)^{-1}=N^{-1}M^{-1}<br />

    N^{-1} is given , you will need to find the inverse of M . Since you haven't been taught on finding its inverse , you shouldn't be doing this question . Do you know how to use the calculator to get its inverse ?
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    hi

     <br />
(MN)^{-1}=N^{-1}M^{-1}<br />

    N^{-1} is given , you will need to find the inverse of M . Since you haven't been taught on finding its inverse , you shouldn't be doing this question . Do you know how to use the calculator to get its inverse ?
    Thanks a lot for the helpful formula...
    The only problem is that this question is one our teacher has given us, meaning we SHOULD be able to solve it...without a calculator. And I'm actually not supposed to use the formula you said either...(just realised the question states this)

    Isn't there some way I could do that?
    Last edited by HelenaStage; January 14th 2010 at 06:56 AM.
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  4. #4
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    Lexington, MA (USA)
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    HGello, Helena!

    Given: . M \:=\:\begin{pmatrix}3&0&8 \\ 0&2&0 \\ 1&0&4 \end{pmatrix}\qquad N^{-1} \;=\;<br />
\begin{pmatrix}1 & \text{-}2 & 3 \\ 3&k&0 \\ 2&4&k \end{pmatrix}

    Find (MN)^{-1}

    Now we have NOT learned how to do the inverse of 3x3 matrices yet, . (Really?)
    so there must be some other way of solving this ... but how?
    If that's true, we can still solve it.

    We know: . (MN)^{-1} \;=\;N^{-1}M^{-1} . . . and we need M^{-1}.


    We can find it the old-fashioned way . . .

    Let: . M^{-1} \:=\:\begin{pmatrix}a&b&c\\d&e&f \\ g&h&i\end{pmatrix}


    Then: . \begin{pmatrix}3&0&8\\0&2&0\\1&0&4\end{pmatrix} \begin{pmatrix}a&b&c\\d&e&f\\g&h&1\end{pmatrix} \;=\;\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatri  x}


    Hence: . \begin{pmatrix}3a+8g & 3b + 8h & 3c+ 8i \\ 2d & 2e & 2f \\ a+4g & b+4h & c + 4i\end{pmatrix} \;=\;\begin{pmatrix}1&0&0 \\ 0&1&0\\0&0&1 \end{pmatrix}


    And we have a system of equations: . \begin{array}{ccccc} 3a+8g \:=\:1 && 3b+8h \:=\:0 && 3c + 8i \:=\:0 \\<br />
2d \:=\:0 && 2e \:=\:1 && 2f \:=\:0 \\ a+4g \:=\:0 && b+4h \:=\:0 && c+4i \:=\:1 \end{array}


    . . which we can solve: . \begin{Bmatrix}a \:=\:1 & b\:=\:0 & c\:=\:\text{-}2 \\  \\[-4mm]<br />
d\:=\:0 & e\:=\:\frac{1}{2} & f \:=\:0 \\ \\[-4mm]<br />
g\:=\:\text{-}\frac{1}{4} & h\:=\:0 & i \:=\:\frac{3}{4} \end{Bmatrix}


    And we have: . M^{-1} \;=\;\begin{pmatrix}<br />
1 & 0 & \text{-}2 \\ \\[-4mm] 0 & \frac{1}{2} & 0 \\ \\[-3mm] \text{-}\frac{1}{4} & 0 & \frac{3}{4} \end{pmatrix}

    Got it?

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  5. #5
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    Quote Originally Posted by Soroban View Post
    HGello, Helena!

    If that's true, we can still solve it.

    We know: . (MN)^{-1} \;=\;N^{-1}M^{-1} . . . and we need M^{-1}.


    We can find it the old-fashioned way . . .

    Let: . M^{-1} \:=\:\begin{pmatrix}a&b&c\\d&e&f \\ g&h&i\end{pmatrix}


    Then: . \begin{pmatrix}3&0&8\\0&2&0\\1&0&4\end{pmatrix} \begin{pmatrix}a&b&c\\d&e&f\\g&h&1\end{pmatrix} \;=\;\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatri  x}


    Hence: . \begin{pmatrix}3a+8g & 3b + 8h & 3c+ 8i \\ 2d & 2e & 2f \\ a+4g & b+4h & c + 4i\end{pmatrix} \;=\;\begin{pmatrix}1&0&0 \\ 0&1&0\\0&0&1 \end{pmatrix}


    And we have a system of equations: . \begin{array}{ccccc} 3a+8g \:=\:1 && 3b+8h \:=\:0 && 3c + 8i \:=\:0 \\<br />
2d \:=\:0 && 2e \:=\:1 && 2f \:=\:0 \\ a+4g \:=\:0 && b+4h \:=\:0 && c+4i \:=\:1 \end{array}


    . . which we can solve: . \begin{Bmatrix}a \:=\:1 & b\:=\:0 & c\:=\:\text{-}2 \\  \\[-4mm]<br />
d\:=\:0 & e\:=\:\frac{1}{2} & f \:=\:0 \\ \\[-4mm]<br />
g\:=\:\text{-}\frac{1}{4} & h\:=\:0 & i \:=\:\frac{3}{4} \end{Bmatrix}


    And we have: . M^{-1} \;=\;\begin{pmatrix}<br />
1 & 0 & \text{-}2 \\ \\[-4mm] 0 & \frac{1}{2} & 0 \\ \\[-3mm] \text{-}\frac{1}{4} & 0 & \frac{3}{4} \end{pmatrix}

    Got it?

    I do! Thanks a lot. So there is actually no other way of finding (MN)^-1 other than using that formula?
    Last edited by HelenaStage; January 14th 2010 at 08:24 AM.
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