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Math Help - Matrix - Gaussian elimination

  1. #1
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    Matrix - Gaussian elimination

    Hi

    Having trouble with solving the equations using Gaussian elimination

    1) Solve the following system of equations using Gaussian elimination

    5x-2y+z=-6
    -2x+z=-4
    -x-2y+2z=3

    Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.

    This is what i have done but it is incorrect according the book's answer.

    \begin{bmatrix}<br />
5  & -2 & 1 &  | &   -6 \\<br />
-2 & 0 & 1 &  | &  -4 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />


    \begin{bmatrix}<br />
5 & -2 & 1 &  | & -6 \\ <br />
0 & -4 & -3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    R2-2R3

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & -4 & 3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    \frac{1}{5}R1

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & -4 & 3 &  | &  -10 \\<br />
0 & \frac{-12}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
 \end{bmatrix}<br />
    R1+R3

    \begin{bmatrix}<br />
 1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
 0 & -4 & 3 &  | &  -10 \\<br />
0 & \frac{-10}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
  \end{bmatrix}<br />
    \frac{1}{4}R2

    \begin{bmatrix}<br />
  1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
  0 & -4 & 3 &  | &  -10 \\<br />
 0 & 0 & \frac{7}{10} &  | &  \frac{-16}{5} \\<br />
   \end{bmatrix}<br />
    \frac{10}{5}R2+R3

    P.S
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  2. #2
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    Would the answer perhaps be:

    x = 2-t
    y = 1/2t - 8
    z = 2t

    If not, then I should take another look at the question If it is, or close, I can post my solution.
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi

    Having trouble with solving the equations using Gaussian elimination

    1) Solve the following system of equations using Gaussian elimination

    5x-2y+z=-6
    -2x+z=-4
    -x-2y+2z=3

    Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.

    This is what i have done but it is incorrect according the book's answer.

    \begin{bmatrix}<br />
5  & -2 & 1 &  | &   -6 \\<br />
-2 & 0 & 1 &  | &  -4 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />


    \begin{bmatrix}<br />
5 & -2 & 1 &  | & -6 \\ <br />
0 & -4 & -3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    R2-2R3

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & -4 & 3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    \frac{1}{5}R1

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & -4 & 3 &  | &  -10 \\<br />
0 & \frac{-12}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
 \end{bmatrix}<br />
    R1+R3

    \begin{bmatrix}<br />
 1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
 0 & -4 & 3 &  | &  -10 \\<br />
0 & \frac{-10}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
  \end{bmatrix}<br />
    \frac{1}{4}R2

    \begin{bmatrix}<br />
  1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
  0 & -4 & 3 &  | &  -10 \\<br />
 0 & 0 & \frac{7}{10} &  | &  \frac{-16}{5} \\<br />
   \end{bmatrix}<br />
    \frac{10}{5}R2+R3

    P.S
    That will give you the correct values for x, y, and z (assuming your arithmetic is correct, I did not check it) but what, exactly, does your text consider the "final tableau"? You have got to "triangular form" and now you can solve for x, y, and z by back substitution. But I notice you got a "1" in the first column but did not get "1"s on the diagonal in the other columns. Some texts, at least, would consider having "1" s all along the diagonal as necessary. Others might expect to see a complete reduction to the identity matrix for the first 3 columns.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    That will give you the correct values for x, y, and z (assuming your arithmetic is correct, I did not check it) but what, exactly, does your text consider the "final tableau"? You have got to "triangular form" and now you can solve for x, y, and z by back substitution. But I notice you got a "1" in the first column but did not get "1"s on the diagonal in the other columns. Some texts, at least, would consider having "1" s all along the diagonal as necessary. Others might expect to see a complete reduction to the identity matrix for the first 3 columns.
    actually the answer is not supplied for this question, but when i used the calculator it gave me what you said '1' on the diagonal

    \begin{bmatrix}<br />
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\<br />
0 & 1 & \frac{-1}{12} & | & \frac{-3}{4} \\<br />
0 & 0 & 1 & | & \frac{-21}{4} \\<br />
\end{bmatrix}
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    Hi

    Having trouble with solving the equations using Gaussian elimination

    1) Solve the following system of equations using Gaussian elimination

    5x-2y+z=-6
    -2x+z=-4
    -x-2y+2z=3

    Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.

    This is what i have done but it is incorrect according the book's answer.

    \begin{bmatrix}<br />
5  & -2 & 1 &  | &   -6 \\<br />
-2 & 0 & 1 &  | &  -4 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />


    \begin{bmatrix}<br />
5 & -2 & 1 &  | & -6 \\ <br />
0 & -4 & -3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    R2-2R3
    There is an error in this first step. -2 times the second number in R3 is + 4. The second number in R2 now should be 4, not -4.

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & -4 & 3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    \frac{1}{5}R1

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & -4 & 3 &  | &  -10 \\<br />
0 & \frac{-12}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
 \end{bmatrix}<br />
    R1+R3

    \begin{bmatrix}<br />
 1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
 0 & -4 & 3 &  | &  -10 \\<br />
0 & \frac{-10}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
  \end{bmatrix}<br />
    \frac{1}{4}R2

    \begin{bmatrix}<br />
  1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
  0 & -4 & 3 &  | &  -10 \\<br />
 0 & 0 & \frac{7}{10} &  | &  \frac{-16}{5} \\<br />
   \end{bmatrix}<br />
    \frac{10}{5}R2+R3

    P.S
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  6. #6
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    ok, i have tried again, this is what i got:

    \begin{bmatrix}<br />
5  & -2 & 1 &  | &   -6 \\<br />
-2 & 0 & 1 &  | &  -4 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />

    \begin{bmatrix}<br />
5 & -2 & 1 &  | & -6 \\ <br />
0 & 4 & -3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    R2-2R3

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & 4 & -3 &  | &  -10 \\<br />
-1 & -2 & 2 &  | &  3 \\<br />
\end{bmatrix}<br />
    \frac{1}{5}R1

    \begin{bmatrix}<br />
 1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
 0 & 4 & -3 &  | &  -10 \\<br />
0 & \frac{-12}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
  \end{bmatrix}<br />
    R3+R1

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & 4 & -3 &  | &  -10 \\<br />
0 & 0 & \frac{2}{10} &  | &  \frac{-21}{5} \\<br />
\end{bmatrix}<br />
    R3+\frac{3}{5}R2

    z=\frac{-105}{10}
    y=\frac{-10-\frac{315}{10}}{10} = \frac{-415}{40}
    x=\frac{-53}{10}
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  7. #7
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    I'm not sure what you are doing, but as I understood it you just have to set Z to t, in my case I set it to 2t to make it more simple counting.

    Then it should all fall out, if z = 2t you can find out what x is and then ultimately what y is.
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  8. #8
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    Quote Originally Posted by Paymemoney View Post
    o
    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & 4 & -3 &  | &  -10 \\<br />
0 & 0 & \frac{2}{10} &  | &  \frac{-21}{5} \\<br />
\end{bmatrix}<br />
    R3+\frac{3}{5}R2

    z=\frac{-105}{10}
    y=\frac{-10-\frac{315}{10}}{10} = \frac{-415}{40}
    x=\frac{-53}{10}
    Briefly looking at that, you should have a figure of z = -210/10 and not -105/10. Then substitute that value into your other equations to work out the value of y and x.
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  9. #9
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    Quote Originally Posted by Paymemoney View Post
    \begin{bmatrix}<br />
 1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
 0 & 4 & -3 &  | &  -10 \\<br />
0 & \frac{-12}{5} & \frac{11}{5} &  | &  \frac{9}{5} \\<br />
  \end{bmatrix}<br />
    R3+R1

    \begin{bmatrix}<br />
1  & \frac{-2}{5} & \frac{1}{5} &  | &   \frac{-6}{5} \\<br />
0 & 4 & -3 &  | &  -10 \\<br />
0 & 0 & \frac{2}{10} &  | &  \frac{-21}{5} \\<br />
\end{bmatrix}<br />
    R3+\frac{3}{5}R2

    z=\frac{-105}{10}
    y=\frac{-10-\frac{315}{10}}{10} = \frac{-415}{40}
    x=\frac{-53}{10}
    Everything looks fine until the 3rd column in your final matrix \frac{3(-3)}{5} = \frac{-9}{5}

    So \frac{11}{5} + \frac{-9}{5} = \frac{2}{5}

    Take \frac{2}{5}R3 and the bottom row becomes 0 0 1 | \frac{-21}{2}
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