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Math Help - Gauss-Jordan complete elimination

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    Gauss-Jordan complete elimination

    Use Gauss-Jordan complete elimination to solve this linear system in four variables.

    \begin{bmatrix}<br />
0&0&1&0\\<br />
0&1&0&0\\<br />
1&0&0&2\\<br />
0&0&2&0\end{bmatrix}

    I've tried this question but can't get really far ... can anyone show a few steps on how to tackle this question
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    Quote Originally Posted by tsal15 View Post
    Use Gauss-Jordan complete elimination to solve this linear system in four variables.

    \begin{bmatrix}<br />
0&0&1&0\\<br />
0&1&0&0\\<br />
1&0&0&2\\<br />
0&0&2&0\end{bmatrix}

    I've tried this question but can't get really far ... can anyone show a few steps on how to tackle this question
    If you have an equation A\mathbf{x} = \mathbf{b} where A is a square matrix and x and b are column vectors, you should be able to solve the system of equations.

    But you haven't given x or b...
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    Quote Originally Posted by Prove It View Post
    If you have an equation A\mathbf{x} = \mathbf{b} where A is a square matrix and x and b are column vectors, you should be able to solve the system of equations.

    But you haven't given x or b...
    oh yes yes you're absolutely correct...i haven't given you guys an x or b...i will now sorry
    b = \begin{bmatrix}16\\4\\15\\0\end{bmatrix}
    x = \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}

    but um i've tried using row operations...still i can't get to a final answer

    Thanks for your continued assistance
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    Quote Originally Posted by tsal15 View Post
    oh yes yes you're absolutely correct...i haven't given you guys an x or b...i will now sorry
    b = \begin{bmatrix}16\\4\\15\\0\end{bmatrix}
    x = \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}

    but um i've tried using row operations...still i can't get to a final answer

    Thanks for your continued assistance
    This system of equations is nonsense.

    Multiply A by x you get...

    A\mathbf{x}=\begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix}.

    So you have \begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix} = \begin{bmatrix}16\\4\\15\\0\end{bmatrix}.

    How can x_3 = 16 AND 0???
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    Quote Originally Posted by Prove It View Post
    This system of equations is nonsense.

    Multiply A by x you get...

    A\mathbf{x}=\begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix}.

    So you have \begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix} = \begin{bmatrix}16\\4\\15\\0\end{bmatrix}.

    How can x_3 = 16 AND 0???
    Do you mean that the matrix has infinitely many solutions? also how did you get your answer? also why did you not choose the method of reducing the matrix to a row echelon form? and how can you solve for x_1 and x_4 if x_2 nor x_3 is part of that equation?

    Thanks again for your continued support
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    Quote Originally Posted by tsal15 View Post
    Do you mean that the matrix has infinitely many solutions? also how did you get your answer? also why did you not choose the method of reducing the matrix to a row echelon form? and how can you solve for x_1 and x_4 if x_2 nor x_3 is part of that equation?

    Thanks again for your continued support
    This is a system of the form A\mathbf{x} = \mathbf{b}.

    I could see that there were a lot of 0's in A, so just by using matrix multiplication, I could multiply A by x and get a vector that is equal to b.

    This system is not only unsolvable, it's nonsense. x_1 and x_4 can not be solved, and x_3 has to equal two different things at once in order for the system to make any sense at all.

    Are you sure you copied the matrix A down correctly?
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    Quote Originally Posted by Prove It View Post
    This is a system of the form A\mathbf{x} = \mathbf{b}.

    I could see that there were a lot of 0's in A, so just by using matrix multiplication, I could multiply A by x and get a vector that is equal to b.

    This system is not only unsolvable, it's nonsense. x_1 and x_4 can not be solved, and x_3 has to equal two different things at once in order for the system to make any sense at all.

    Are you sure you copied the matrix A down correctly?
    Oh my god you are absolutely correct... that matrix was for another question sorry to have wasted your time but this is the true matrix A:

    \begin{bmatrix}1&2&4&8\\1&-1&-2&2\\0&3&6&9\\-1&2&4&0\end{bmatrix}

    Now, i've checked and double checked and so far this matrix A is correctly copied. also, x and b still are the same.

    Sorry to have caused such trouble
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    Quote Originally Posted by Prove It View Post
    This is a system of the form A\mathbf{x} = \mathbf{b}.

    I could see that there were a lot of 0's in A, so just by using matrix multiplication, I could multiply A by x and get a vector that is equal to b.

    This system is not only unsolvable, it's nonsense. x_1 and x_4 can not be solved, and x_3 has to equal two different things at once in order for the system to make any sense at all.

    Are you sure you copied the matrix A down correctly?
    Sorry but can you still help? thanks Prove It
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    \left[ \begin{array}{cccc|c}<br />
1 & 2 & 4 & 8 & 16\\<br />
1 & -1 & -2 & 2 & 4\\<br />
0 & 3 & 6 & 9 & 15\\<br />
-1 & 2 & 4 & 0 & 0\end{array}\right]

    Do [tex]R_2 - R_1[tex] and R_4 + R_1

    \left[ \begin{array}{cccc|c}<br />
1 & 2 & 4 & 8 & 16\\<br />
0 & -3 & -6 & -6 & -12\\<br />
0 & 3 & 6 & 9 & 15\\<br />
0 & 4 & 8 & 8 & 16\end{array}\right]

    Do -\frac{1}{3}R_2

    \left[ \begin{array}{cccc|c}<br />
 1 & 2 & 4 & 8 & 16\\<br />
0 & 1 & 2 & 2 & 4\\<br />
 0 & 3 & 6 & 9 & 15\\<br />
0 & 4 & 8 & 8 & 16\end{array}\right]

    Do R_3 - 3R_2 and R_4 - 4R_2

    \left[ \begin{array}{cccc|c}<br />
  1 & 2 & 4 & 8 & 16\\<br />
 0 & 1 & 2 & 2 & 4\\<br />
  0 & 0 & 0 & 3 & 3\\<br />
 0 & 0 & 0 & 0 & 0\end{array}\right].

    So we have...

    3x_4 = 3 so x_4 = 1.

    Notice that we can not solve for x_3. So we call x_3 = t, an arbitrary parameter.

    Back substituting gives

    x_2 + 2x_3 + 2x_4 = 4

    x_2 + 2t + 2(1) = 4

    x_2 = 2 - 2t


    x_1 + 2x_2 + 4x_3 + 8x_4 = 16

    x_1 + 2(2 - 2t) + 4t + 8(1) = 16

    x_1 + 4 - 4t + 4t + 8 = 16

    x_1 = 4


    So \mathbf{x} = \left[ \begin{array}{c}<br />
1\\<br />
 2 - 2t\\<br />
t\\<br />
 1\end{array}\right].
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    Quote Originally Posted by Prove It View Post
    \left[ \begin{array}{cccc|c}<br />
1 & 2 & 4 & 8 & 16\\<br />
1 & -1 & -2 & 2 & 4\\<br />
0 & 3 & 6 & 9 & 15\\<br />
-1 & 2 & 4 & 0 & 0\end{array}\right]

    Do [tex]R_2 - R_1[tex] and R_4 + R_1

    \left[ \begin{array}{cccc|c}<br />
1 & 2 & 4 & 8 & 16\\<br />
0 & -3 & -6 & -6 & -12\\<br />
0 & 3 & 6 & 9 & 15\\<br />
0 & 4 & 8 & 8 & 16\end{array}\right]

    Do -\frac{1}{3}R_2

    \left[ \begin{array}{cccc|c}<br />
 1 & 2 & 4 & 8 & 16\\<br />
0 & 1 & 2 & 2 & 4\\<br />
 0 & 3 & 6 & 9 & 15\\<br />
0 & 4 & 8 & 8 & 16\end{array}\right]

    Do R_3 - 3R_2 and R_4 - 4R_2

    \left[ \begin{array}{cccc|c}<br />
  1 & 2 & 4 & 8 & 16\\<br />
 0 & 1 & 2 & 2 & 4\\<br />
  0 & 0 & 0 & 3 & 3\\<br />
 0 & 0 & 0 & 0 & 0\end{array}\right].

    So we have...

    3x_4 = 3 so x_4 = 1.

    Notice that we can not solve for x_3. So we call x_3 = t, an arbitrary parameter.

    Back substituting gives

    x_2 + 2x_3 + 2x_4 = 4

    x_2 + 2t + 2(1) = 4

    x_2 = 2 - 2t


    x_1 + 2x_2 + 4x_3 + 8x_4 = 16

    x_1 + 2(2 - 2t) + 4t + 8(1) = 16

    x_1 + 4 - 4t + 4t + 8 = 16

    x_1 = 4


    So \mathbf{x} = \left[ \begin{array}{c}<br />
1\\<br />
 2 - 2t\\<br />
t\\<br />
 1\end{array}\right].
    hey, that is sooooo helpful, if only my textbook could be as helpful as you are thanks for your help Prove it
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    Quote Originally Posted by Prove It View Post
    \left[ \begin{array}{cccc|c}<br />
1 & 2 & 4 & 8 & 16\\<br />
1 & -1 & -2 & 2 & 4\\<br />
0 & 3 & 6 & 9 & 15\\<br />
-1 & 2 & 4 & 0 & 0\end{array}\right]

    Do [tex]R_2 - R_1[tex] and R_4 + R_1

    \left[ \begin{array}{cccc|c}<br />
1 & 2 & 4 & 8 & 16\\<br />
0 & -3 & -6 & -6 & -12\\<br />
0 & 3 & 6 & 9 & 15\\<br />
0 & 4 & 8 & 8 & 16\end{array}\right]

    Do -\frac{1}{3}R_2

    \left[ \begin{array}{cccc|c}<br />
 1 & 2 & 4 & 8 & 16\\<br />
0 & 1 & 2 & 2 & 4\\<br />
 0 & 3 & 6 & 9 & 15\\<br />
0 & 4 & 8 & 8 & 16\end{array}\right]

    Do R_3 - 3R_2 and R_4 - 4R_2

    \left[ \begin{array}{cccc|c}<br />
  1 & 2 & 4 & 8 & 16\\<br />
 0 & 1 & 2 & 2 & 4\\<br />
  0 & 0 & 0 & 3 & 3\\<br />
 0 & 0 & 0 & 0 & 0\end{array}\right].

    So we have...

    3x_4 = 3 so x_4 = 1.

    Notice that we can not solve for x_3. So we call x_3 = t, an arbitrary parameter.

    Back substituting gives

    x_2 + 2x_3 + 2x_4 = 4

    x_2 + 2t + 2(1) = 4

    x_2 = 2 - 2t


    x_1 + 2x_2 + 4x_3 + 8x_4 = 16

    x_1 + 2(2 - 2t) + 4t + 8(1) = 16

    x_1 + 4 - 4t + 4t + 8 = 16

    x_1 = 4


    So \mathbf{x} = \left[ \begin{array}{c}<br />
1\\<br />
 2 - 2t\\<br />
t\\<br />
 1\end{array}\right].
    I did all the reduction parts but that was where i got stuck now i got something i can learn from So, how did you know we cannot solve for x_3? that is what has been holding me up from completing this question.

    Thanks again
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    Quote Originally Posted by tsal15 View Post
    I did all the reduction parts but that was where i got stuck now i got something i can learn from So, how did you know we cannot solve for x_3? that is what has been holding me up from completing this question.

    Thanks again
    Because there isn't an equation we can back-substitute x_4 into that can give us JUST x_3.

    When we back substitute, we get an equation that has both x_2 and x_3.

    So the best we can do is write one of them in terms of the other.

    In other words, we can let x_3 be whatever we like, it won't change the system. The only thing that will change is x_2, because it depends on x_3.
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    Quote Originally Posted by Prove It View Post
    Because there isn't an equation we can back-substitute x_4 into that can give us JUST x_3.

    When we back substitute, we get an equation that has both x_2 and x_3.

    So the best we can do is write one of them in terms of the other.

    In other words, we can let x_3 be whatever we like, it won't change the system. The only thing that will change is x_2, because it depends on x_3.
    uhuh so does that mean we could've let x_2=t?
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    Quote Originally Posted by tsal15 View Post
    uhuh so does that mean we could've let x_2=t?
    Yep. Either of them can be kept arbitrary. We just take note that one depends on the other.
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    Quote Originally Posted by Prove It View Post
    Yep. Either of them can be kept arbitrary. We just take note that one depends on the other.
    You have been very helpful thank you

    I don't know if i'm asking for too much here, but i've got another thread "Gaussian - Jordan Elimination" - would you be able to have some input on that? thanks again Prove It
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