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Solving Linear Equations (Matrices)

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that the equation

    [itex]\left(
    \begin{array}{ccc}
    1 & 2 & -3 \\
    2 & 6 & -11 \\
    1 & -2 & 7
    \end{array}
    \right)\left(
    \begin{array}{c}
    x \\
    y \\
    z
    \end{array}
    \right)=\left(
    \begin{array}{c}
    a \\
    b \\
    c
    \end{array}
    \right)[/itex]

    Is only soluble if [itex]c+2b-5a=0[/itex]

    (b) Hence show the planes


    [itex]x+2y-3z=1[/itex]

    [itex]2x+6y-11z=2[/itex]

    [itex]x-2y+7z=1[/itex]

    Intersect in a line.

    (c) Find in terms of s, the co-ordinate of the point in whihc this line meets the plane z=s.

    3. The attempt at a solution
    [itex]
    \text{Det}\left[\left(
    \begin{array}{ccc}
    1 & 2 & -3 \\
    2 & 6 & -11 \\
    1 & -2 & 7
    \end{array}
    \right)\right]=0[/itex]

    [itex]x+2y-3z=a[/itex]

    [itex]2x+6y-11z=b[/itex]

    [itex]x-2y+7z=c[/itex]

    [itex]c+2b-5a = x-2y+7z + 2(2x+6y-11z) -5(x+2y-3z)=0[/itex]

    Not sure whether this is sufficient?

    (b)

    [itex]2x+4y-6z=2[/itex]

    [itex]2x+6y-11z=2[/itex]

    [itex]y=\frac{5}{2}z[/itex]

    [itex]x=1-2z[/itex]

    [itex] z=\lambda[/itex]

    [itex] x=1-2\lambda, y=\frac{5}{2}\lambda[/itex]

    What is the vector equation of this result if it is correct?

    [itex] r = \left(
    \begin{array}{c}
    1 \\
    0 \\
    0
    \end{array}
    \right) + \lambda\left(
    \begin{array}{c}
    -2 \\
    \frac{5}{2} \\
    1
    \end{array}
    \right)[/itex]

    Is that right?

    (c) Substitute z=s

    [itex](1-2s,\frac{5s}{2},s)[/itex]
     
  2. jcsd
  3. Jun 16, 2009 #2

    statdad

    User Avatar
    Homework Helper

    You could try reducing the augmented matrix to RREF and looking at what must happen to have a solution.

    [tex]
    \begin{pmatrix}
    1 & 2 & -3 & a\\
    2 & 6 & -11 & b \\
    1 & -2 & 7 & c
    \end{pmatrix}
    [/tex]
     
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