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Math Help - Solving a system of linear equations using matrices.

  1. #1
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    Post Solving a system of linear equations using matrices.

    the cartesian equations of 3 planes are:
    x + y + z = 3
    3x - y - z = 1
    x + py = q

    By writing down the augmented matrix for this system and reducing it to echelon form,
    find for what values of p and q these planes
    (a) intersect at a single point;
    (b) intersect along a line;
    (c) have no common line or point of intersection.
    In case (a) find, in terms of p and q, the coordinates of the point, and in case (b) find the scalar parametric equations of the line.

    i got the row echelon form to be
    1 2 2 | 3
    0 1 2 | 2
    0 0 2p | q+1+2p

    is the above true and if so are these true?
    case A)
    z = (q+1+2p)/2p
    y=2-(2q+2+4p)/2p
    x=1

    case B) DONT UNDERSTAND

    case C) at p = 0 ; q diff than -1

    Im soooo confused.. ALL HELP APPRECIATED!
    Last edited by mr fantastic; May 11th 2010 at 06:50 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by swiftshift View Post
    the cartesian equations of 3 planes are:
    x + y + z = 3
    3x - y - z = 1
    x + py = q

    By writing down the augmented matrix for this system and reducing it to echelon form,
    find for what values of p and q these planes
    (a) intersect at a single point;
    (b) intersect along a line;
    (c) have no common line or point of intersection.
    In case (a) find, in terms of p and q, the coordinates of the point, and in case (b) find the scalar parametric equations of the line.

    i got the row echelon form to be
    1 2 2 | 3
    0 1 2 | 2
    0 0 2p | q+1+2p

    is the above true and if so are these true?
    case A)
    z = (q+1+2p)/2p
    y=2-(2q+2+4p)/2p
    x=1

    case B) DONT UNDERSTAND

    case C) at p = 0 ; q diff than -1

    Im soooo confused.. ALL HELP APPRECIATED!
    \begin{bmatrix}<br />
1 & 1 & 1 & 3\\ <br />
3 & -1 & -1 & 1\\ <br />
1 & p & 0 & q<br />
\end{bmatrix}

    ref=\begin{bmatrix}<br />
1 & 1 & 1 & 3\\ <br />
0 & 1 & 1 & 2\\ <br />
0 & 0 & p & 2p+1-q<br />
\end{bmatrix}

    Part a is when p=1 and q\in\mathbb{R}
    Part b (still thinking)
    Part c is when p=0 and q\in\mathbb{R}, q\neq1
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  3. #3
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    For any system of equations, there are 3 possibilities: there is a unique solution, there is NO solution, or there exist an infinite number of solutions.

    Your row reduce matrix is \begin{bmatrix}<br />
1 & 1 & 1 & 3\\ <br />
0 & 1 & 1 & 2\\ <br />
0 & 0 & p & 2p+1-q<br />
\end{bmatrix}
    which is equivalent to the equations x+ y+ z= 3, y+ z= 2, and pz= 2p+1- q.

    As long as p\ne 0, we can divide both sides of the last equation by p to get z= \frac{2p+ 1- q}{p}, then use that value in y+ z= 2 to solve for y, then use those in x+ y+ z= 3 to solve for x.

    If p= 0 and 2p+ 1- q= 1- q\ne 0, then the last equation is
    0z= 2p+ 1- q which is impossible- 0 times any number is 0. There is no solution, the three planes do not intersect.

    If p= 0 and 2p+ 1- q= 1- q= 0[/tex], that is, if p= 0 and q= 1, the last equation is 0z= 0 which is true for all z. In that case, the second equation becomes y+ z= 2 or y= 2- z. The first equation, x+ y+ z= 3 becomes x+ 2= 3 so x= 1. (1, 2- z, z) for any z will satisfy the equations. Any point along that line (in parametric equations, x= 1, y= 2- t, z= t) will satisfy those equations so the three planes intersect on that line.
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