0
$\begingroup$

$\begin{pmatrix} -1&3&2&1\\ 4&-12&-7&-5 \\ 3&-9&-4&-5 \end{pmatrix}\begin{pmatrix} x_1\\x_2\\x_3\\x_4 \end{pmatrix} = \begin{pmatrix} 1\\0\\5\end{pmatrix}$

My attempt:

Reduce it to reduced Echolon form: $\begin{pmatrix} 1& -3& 0& -3& 7&\\ 0& 0& 1& -1& 4&\\ 0& 0& 0& 0& 0&\\\end{pmatrix}$

Now take the standard solution+the characteristic columns to form the homogenous solutionspace:

$\begin{pmatrix} 7\\4\\0\\0\end{pmatrix}+ lin\begin{pmatrix} \begin{pmatrix} 3\\1\\0\\0\end{pmatrix},\begin{pmatrix} 3\\1\\0\\1\end{pmatrix}\end{pmatrix}$

Now when I multiply A with $\begin{pmatrix} 7\\4\\0\\0\end{pmatrix}+\begin{pmatrix} 3\\1\\0\\0\end{pmatrix} = \begin{pmatrix} 10\\5\\0\\0\end{pmatrix}$, it should give me $\begin{pmatrix} 1\\0\\5\end{pmatrix}$, but it is $\begin{pmatrix} 5\\-20\\-15\end{pmatrix}$

Why does this happen? What is my missconception in here?

EDIT: The standard solution should be $\begin{pmatrix} 7\\0\\4\\0\end{pmatrix}$ instead

  • 2
    From your RRE form, the 'standard solution' should be $(7,0,4,0)^{T}$ not $(7,4,0,0)^{T}$. You have no leading $1$ for $x_{2}$ and $x_{4}$. Also, the 'characteristic columns' should be $(3,1,0,0)^{T}$ and $(3,0,1,1)^{T}$. – 2017-01-30
  • 0
    @Mattos Thanks a lot, dumb little mistakes.. – 2017-01-30

0 Answers 0