Basis perpendicular to a subspace spanned by 2 vectors

**zzzoak** · April 19th 2010, 03:16 PM

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.

**dwsmith** · April 19th 2010, 04:26 PM

Do the rref of this matrix

$\begin{bmatrix} 1 & -4 & 1 & 4 & 0\\ 3 & -13 & 3 & -4 & 0 \end{bmatrix}\rightarrow\begin{bmatrix} 1 & 0 & 1 & 68 & 0\\ 0 & 1 & 0 & 16 & 0 \end{bmatrix}$

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

$\begin{bmatrix} -x_3-68x_4\\ -16x_4\\ x_3\\ x_4 \end{bmatrix}\rightarrow x_3\begin{bmatrix} -1\\ 0\\ 1\\ 0 \end{bmatrix}+x_4\begin{bmatrix} -68\\ -16\\ 0\\ 1 \end{bmatrix}$

**mybrohshi5** · April 19th 2010, 06:06 PM

Originally Posted by dwsmith

Do the rref of this matrix

$\begin{bmatrix} 1 & -4 & 1 & 4 & 0\\ 3 & -13 & 3 & -4 & 0 \end{bmatrix}\rightarrow\begin{bmatrix} 1 & 0 & 1 & 68 & 0\\ 0 & 1 & 0 & 16 & 0 \end{bmatrix}$

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

$\begin{bmatrix} -x_3-68x_4\\ -16x_4\\ x_3\\ x_4 \end{bmatrix}\rightarrow x_3\begin{bmatrix} -1\\ 0\\ 1\\ 0 \end{bmatrix}+x_4\begin{bmatrix} -68\\ -16\\ 0\\ 1 \end{bmatrix}$

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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