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Math Help - [SOLVED] using matrices to solve for a system of equations using gauss-jordan elimina

  1. #1
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    [SOLVED] using matrices to solve for a system of equations using gauss-jordan elimina

    i have to use matrices to solve the system of equations using gauss-jordan elimination

    my system of equations is:
    2x -y +3z =24
    2y -z =14
    7x -5y =6

    my matrix would be
    2 -1 3 24
    0 2 -1 14
    7 -5 0 6

    can i have a walk through of this equations and an explanation on what to do? this is what i have so far:

    i want to get this matrix into reduced row-echelon form, so first i put it into row-echelon form then reduce.
    -i times Row 1 by 1/2 so i would end up with:
    1 -1/2 3/2 12
    0 2 -1 14
    7 -5 0 6

    -then i times Row 2 by 1/2 and end up with:
    1 -1/2 3/2 12
    0 1 -1/2 7
    7 -5 0 6

    -then i times Row 1 by -7 and add it to Row 3 and end up with:
    1 -1/2 3/2 12
    0 2 -1 14
    0 -1.5 -10.5 -78

    what would i do to get the leading 1 of row 3? am i doing this right? are my answers right so far? can anyone give me a walkthrough through the rest of the problem? sorry, my spacing isn't working.
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  2. #2
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    calculator answer=........ x = 8, y = 10, and z = 6
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  3. #3
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    Quote Originally Posted by somanyquestions View Post
    i have to use matrices to solve the system of equations using gauss-jordan elimination

    my system of equations is:
    2x -y +3z =24
    2y -z =14
    7x -5y =6

    my matrix would be
    2 -1 3 24
    0 2 -1 14
    7 -5 0 6

    can i have a walk through of this equations and an explanation on what to do? this is what i have so far:

    i want to get this matrix into reduced row-echelon form, so first i put it into row-echelon form then reduce.
    -i times Row 1 by 1/2 so i would end up with:
    1 -1/2 3/2 12
    0 2 -1 14
    7 -5 0 6

    -then i times Row 2 by 1/2 and end up with:
    1 -1/2 3/2 12
    0 1 -1/2 7
    7 -5 0 6

    -then i times Row 1 by -7 and add it to Row 3 and end up with:
    1 -1/2 3/2 12
    0 2 -1 14
    0 -1.5 -10.5 -78

    what would i do to get the leading 1 of row 3? am i doing this right? are my answers right so far? can anyone give me a walkthrough through the rest of the problem? sorry, my spacing isn't working.
    \left[\begin{matrix}<br />
2 & -1 & 3 &|& 24\\<br />
0 & 2 & -1 &|& 14\\<br />
7 & -5 & 0 &|& 6 \end{matrix}\right]

    \frac{1}{2}R_1 \to R_1

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 2 & -1 &|& 14\\<br />
7 & -5 & 0 &|& 6 \end{matrix}\right]

    R_3 - 7R_1 \to R_3

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 2 & -1 &|& 14\\<br />
0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]

    \frac{1}{2}R_2 \to R_2

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 1 & -\frac{1}{2} &|& 7\\<br />
0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]

    R_3 + \frac{3}{2}R_2 \to R_3

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 1 & -\frac{1}{2} &|& 7\\<br />
0 & 0 & -\frac{45}{4} &|& -\frac{135}{2} \end{matrix}\right]

    -\frac{4}{45}R_3 \to R_3

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 1 & -\frac{1}{2} &|& 7\\<br />
0 & 0 & 1 &|& 6 \end{matrix}\right]


    Now it is in row-echelon form and you can solve the system of equations.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    \left[\begin{matrix}<br />
2 & -1 & 3 &|& 24\\<br />
0 & 2 & -1 &|& 14\\<br />
7 & -5 & 0 &|& 6 \end{matrix}\right]

    \frac{1}{2}R_1 \to R_1

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 2 & -1 &|& 14\\<br />
7 & -5 & 0 &|& 6 \end{matrix}\right]

    R_3 - 7R_1 \to R_3

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 2 & -1 &|& 14\\<br />
0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]

    \frac{1}{2}R_2 \to R_2

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 1 & -\frac{1}{2} &|& 7\\<br />
0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]

    R_3 + \frac{3}{2}R_2 \to R_3

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 1 & -\frac{1}{2} &|& 7\\<br />
0 & 0 & -\frac{45}{4} &|& -\frac{135}{2} \end{matrix}\right]

    -\frac{4}{45}R_3 \to R_3

    \left[\begin{matrix}<br />
1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\<br />
0 & 1 & -\frac{1}{2} &|& 7\\<br />
0 & 0 & 1 &|& 6 \end{matrix}\right]


    Now it is in row-echelon form and you can solve the system of equations.
    how do you type out all these matrices?
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  5. #5
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    Quote Originally Posted by somanyquestions View Post
    how do you type out all these matrices?
    Click on them - that gives you the code I used.
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  6. #6
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    can you please make your answer into reduced row-echelon form?
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  7. #7
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    Quote Originally Posted by somanyquestions View Post
    can you please make your answer into reduced row-echelon form?
    The complete solution has almost been handed to you. Surely you can do the last few steps to get the reduced row echelon form.
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  8. #8
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    yeah, it's fine i was just freaking out.
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