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Math Help - Basis perpendicular to a subspace spanned by 2 vectors

  1. #16
    Senior Member
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    a-4b+c+4d=0
    3a-13b+3c-4d=0
    1)
    c=0 d=1
    a-4b+4=0
    3a-13b-4=0
    a=-68 b=-16
    2)
    c=1 d=0
    a-4b+1=0
    3a-13b+3=0
    a=-1 b=0.
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  2. #17
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    Do the rref of this matrix

    \begin{bmatrix}<br />
1 & -4 & 1 & 4 & 0\\ <br />
3 & -13 & 3 & -4 & 0<br />
\end{bmatrix}\rightarrow\begin{bmatrix}<br />
1 & 0 & 1 & 68 & 0\\ <br />
0 & 1 & 0 & 16 & 0<br />
\end{bmatrix}

    x_1=-x_3-68x_4
    x_2=-16x_4
    x_3
    x_4

    \begin{bmatrix}<br />
-x_3-68x_4\\ <br />
-16x_4\\ <br />
x_3\\ <br />
x_4<br />
\end{bmatrix}\rightarrow x_3\begin{bmatrix}<br />
-1\\ <br />
0\\ <br />
1\\ <br />
0<br />
\end{bmatrix}+x_4\begin{bmatrix}<br />
-68\\ <br />
-16\\ <br />
0\\ <br />
1<br />
\end{bmatrix}
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  3. #18
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230
    Quote Originally Posted by dwsmith View Post
    Do the rref of this matrix

    \begin{bmatrix}<br />
1 & -4 & 1 & 4 & 0\\ <br />
3 & -13 & 3 & -4 & 0<br />
\end{bmatrix}\rightarrow\begin{bmatrix}<br />
1 & 0 & 1 & 68 & 0\\ <br />
0 & 1 & 0 & 16 & 0<br />
\end{bmatrix}

    x_1=-x_3-68x_4
    x_2=-16x_4
    x_3
    x_4

    \begin{bmatrix}<br />
-x_3-68x_4\\ <br />
-16x_4\\ <br />
x_3\\ <br />
x_4<br />
\end{bmatrix}\rightarrow x_3\begin{bmatrix}<br />
-1\\ <br />
0\\ <br />
1\\ <br />
0<br />
\end{bmatrix}+x_4\begin{bmatrix}<br />
-68\\ <br />
-16\\ <br />
0\\ <br />
1<br />
\end{bmatrix}
    Yeah my professor decided to teach us how to do this today so i got that answer as well when i worked it out

    Thanks for all the help.
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