# Thread: Basis perpendicular to a subspace spanned by 2 vectors

1. 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.

2. Do the rref of this matrix

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

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

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

3. 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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