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Math Help - Matrix(Stucked at part iii)

  1. #1
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    Matrix(Stucked at part iii)

    Question

    i)Find the inverse of  \left[ \begin{array}{cc}<br />
2 & k-2 \\<br />
3 & k \end{array} \right]

    ii) Hence find the solutions of the simultaneous equations
    2x+(k-2)y=4
    3x+ky=5

    Giving your answers in terms of K

    iii) State the value of k for which there are no solutions to the simultaneous equations.

    Attempted solutions

    i) <br />
\left[ \begin{array}{cc}2 & {k-2} \\3 & k \end{array} \right]^{-1} = \frac{1}{2k-3(k-2)}\left[ \begin{array}{cc}k & -(k-2) \\-3 & 2 \end{array} \right]<br />

    <br />
= \frac{1}{2k-(3k-6)}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]<br />

    <br />
= \frac{1}{-k+6}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right] <br />

    <br />
= \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \right]<br />

    ii) \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \left[ \begin{array}{cc}<br />
4 \\<br />
5 \end{array} \right]\right]

     <br />
=\left[ \begin{array}{cc}\frac{4k}{-k+6} + \frac{5(-k+2)}{-k+6} \\ \frac{-12}{-k+6} + \frac{10}{-k+6} \end{array} \right]<br />

     <br />
= \left[ \begin{array}{cc}\frac{-k+10}{-k+6} \\\frac{-2}{-k+6} \end{array} \right]<br />

    x=\frac{-k+10}{-k+6} y=\frac{-2}{-k+6}

    Stucked @ iii
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  2. #2
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    Quote Originally Posted by Punch View Post
    Question

    i)Find the inverse of  \left[ \begin{array}{cc}<br />
2 & k-2 \\<br />
3 & k \end{array} \right]

    ii) Hence find the solutions of the simultaneous equations
    2x+(k-2)y=4
    3x+ky=5

    Giving your answers in terms of K

    iii) State the value of k for which there are no solutions to the simultaneous equations.

    Attempted solutions

    i) <br />
\left[ \begin{array}{cc}2 & {k-2} \\3 & k \end{array} \right]^{-1} = \frac{1}{2k-3(k-2)}\left[ \begin{array}{cc}k & -(k-2) \\-3 & 2 \end{array} \right]<br />

    <br />
= \frac{1}{2k-(3k-6)}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]<br />

    <br />
= \frac{1}{-k+6}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right] <br />

    <br />
= \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \right]<br />

    ii) \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \left[ \begin{array}{cc}<br />
4 \\<br />
5 \end{array} \right]\right]

     <br />
=\left[ \begin{array}{cc}\frac{4k}{-k+6} + \frac{5(-k+2)}{-k+6} \\ \frac{-12}{-k+6} + \frac{10}{-k+6} \end{array} \right]<br />

     <br />
= \left[ \begin{array}{cc}\frac{-k+10}{-k+6} \\\frac{-2}{-k+6} \end{array} \right]<br />

    x=\frac{-k+10}{-k+6} y=\frac{-2}{-k+6}

    Stucked @ iii
    by gausssian elimination, we have (in augmented form)

     \left[ \begin{array}{ccc}<br />
2 & k-2 & 4\\<br />
 0 & \frac{6-k}{2} & 1\end{array} \right]

    this system of equations will have no solution if \frac{6-k}{2} = 0, that is for k = 6
    Last edited by dedust; January 16th 2010 at 10:35 PM.
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  3. #3
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    Quote Originally Posted by dedust View Post
    by gausssian elimination, we have (in augmented form)

     \left[ \begin{array}{ccc}<br />
2 & k-2 & 4\\<br />
0 & \frac{3-k}{2} & 1\end{array} \right]

    this system of equations will have no solution if \frac{3-k}{2} = 0, that is for k = 3
    after going through a few examples in the book, maybe you are wrong and answer to part iii should be k=6

    this is why
    x=\frac{-k+10}{-k+6} y=\frac{-2}{-k+6}
    since denomnator = -k+6, if k=6, -6+6=0
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  4. #4
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    Quote Originally Posted by Punch View Post
    Question

    i)Find the inverse of  \left[ \begin{array}{cc}<br />
2 & k-2 \\<br />
3 & k \end{array} \right]

    ii) Hence find the solutions of the simultaneous equations
    2x+(k-2)y=4
    3x+ky=5

    Giving your answers in terms of K

    iii) State the value of k for which there are no solutions to the simultaneous equations.

    Attempted solutions

    i) <br />
\left[ \begin{array}{cc}2 & {k-2} \\3 & k \end{array} \right]^{-1} = \frac{1}{2k-3(k-2)}\left[ \begin{array}{cc}k & -(k-2) \\-3 & 2 \end{array} \right]<br />

    <br />
= \frac{1}{2k-(3k-6)}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]<br />

    <br />
= \frac{1}{-k+6}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right] <br />

    <br />
= \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \right]<br />

    ii) \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \left[ \begin{array}{cc}<br />
4 \\<br />
5 \end{array} \right]\right]

     <br />
=\left[ \begin{array}{cc}\frac{4k}{-k+6} + \frac{5(-k+2)}{-k+6} \\ \frac{-12}{-k+6} + \frac{10}{-k+6} \end{array} \right]<br />

     <br />
= \left[ \begin{array}{cc}\frac{-k+10}{-k+6} \\\frac{-2}{-k+6} \end{array} \right]<br />

    x=\frac{-k+10}{-k+6} y=\frac{-2}{-k+6}

    Stucked @ iii
    If A\mathbf{x} = \mathbf{b}

    Then \mathbf{x} = A^{-1}\mathbf{b}.


    This will only happen if A^{-1} exists.


    A^{-1} exists if |A| \neq 0.


    So in your case

    |A| = - k + 6.


    What value of k will make |A| = 0?
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  5. #5
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    Quote Originally Posted by Punch View Post
    after going through a few examples in the book, maybe you are wrong and answer to part iii should be k=6

    this is why
    x=\frac{-k+10}{-k+6} y=\frac{-2}{-k+6}
    since denomnator = -k+6, if k=6, -6+6=0
    oops,..wrong in gaussian elimination,..it should be \left[ \begin{array}{ccc}<br />
2 & k-2 & 4\\<br />
 0 & \frac{6-k}{2} & 1\end{array} \right]
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  6. #6
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    Quote Originally Posted by dedust View Post
    oops,..wrong in gaussian elimination,..it should be \left[ \begin{array}{ccc}<br />
2 & k-2 & 4\\<br />
0 & \frac{6-k}{2} & 1\end{array} \right]
    gaussian elimination is alien to me, the simplest method is probably to turn denominator equals to zero
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  7. #7
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    Quote Originally Posted by Punch View Post
    gaussian elimination is alien to me, the simplest method is probably to turn denominator equals to zero
    The denominator you are referring to is called the "determinant".

    The determinant of A is |A|.

    See my post above...
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  8. #8
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    Quote Originally Posted by Prove It View Post
    The denominator you are referring to is called the "determinant".

    The determinant of A is |A|.

    See my post above...
    Yay I got it now, if ad-bc=0 it means that equation has no solution !
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