Assume that I have a fricas (1) -> L := [[- 1,
Type: List(List(Integer))
(which is calculated from some earlier expressions).
How can I convert it into a fricas L2:=[concat([L.i.j for j in 1..#L.i],
Type: List(List(Integer))
Namely I want to have a matrix like: fricas A := matrix L2
Type: Matrix(Integer)
so that I can do fricas vp := vector[p0,
and fricas vp * A
Type: Vector(Polynomial(Integer))
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last edited 17 years ago |
![\label{eq1}\left[{\left[ - 1, \: 3, \: - 3, \: 1 \right]}, \:{\left[ 3, \: - 6, \: 3 \right]}, \:{\left[ - 3, \: 3 \right]}, \:{\left[ 1 \right]}\right]](images/5490416470170246866-16.0px.png "\label{eq1}\left[{\left[ - 1, \: 3, \: - 3, \: 1 \right]}, \:{\left[ 3, \: - 6, \: 3 \right]}, \:{\left[ - 3, \: 3 \right]}, \:{\left[ 1 \right]}\right]")
![\label{eq2}\left[{\left[ - 1, \: 3, \: - 3, \: 1 \right]}, \:{\left[ 3, \: - 6, \: 3, \: 0 \right]}, \:{\left[ - 3, \: 3, \: 0, \: 0 \right]}, \:{\left[ 1, \: 0, \: 0, \: 0 \right]}\right]](images/7793675996418021589-16.0px.png "\label{eq2}\left[{\left[ - 1, \: 3, \: - 3, \: 1 \right]}, \:{\left[ 3, \: - 6, \: 3, \: 0 \right]}, \:{\left[ - 3, \: 3, \: 0, \: 0 \right]}, \:{\left[ 1, \: 0, \: 0, \: 0 \right]}\right]")
![\label{eq3}\left[
\begin{array}{cccc}
- 1 & 3 & - 3 & 1
\
3 & - 6 & 3 & 0
\
- 3 & 3 & 0 & 0
\
1 & 0 & 0 & 0](images/2046356664714734879-16.0px.png "\label{eq3}\left[
\begin{array}{cccc}
- 1 & 3 & - 3 & 1
\
3 & - 6 & 3 & 0
\
- 3 & 3 & 0 & 0
\
1 & 0 & 0 & 0")
![\label{eq4}\left[ p 0, \: p 1, \: p 2, \: p 3 \right]](images/2807874791227082489-16.0px.png "\label{eq4}\left[ p 0, \: p 1, \: p 2, \: p 3 \right]")
![\label{eq5}\left[{p 3 -{3 \ p 2}+{3 \ p 1}- p 0}, \:{{3 \ p 2}-{6 \ p 1}+{3 \ p 0}}, \:{{3 \ p 1}-{3 \ p 0}}, \: p 0 \right]](images/3173458622616540898-16.0px.png "\label{eq5}\left[{p 3 -{3 \ p 2}+{3 \ p 1}- p 0}, \:{{3 \ p 2}-{6 \ p 1}+{3 \ p 0}}, \:{{3 \ p 1}-{3 \ p 0}}, \: p 0 \right]")