How do i fix the error that i am getting?

2 次查看（过去 30 天）
Taylor
Taylor 2023-7-13
评论： Torsten 2023-7-13
My code is
A = [1,1,0,2,0;0,1,1,3,0;2,0,0,0,1;3,1,0,2,1;2,1,1,3,0;1,0,0,2,1]
rrefA = rref(A)
basisRowSpace = A(rrefA(:,end)~=0,:)
pivotColumns = rrefA(:,1:end-1)
basisColumnSpace = A(:,pivotColumns)
This is where it stops working and I dont know how to fix it.
The error that I get is Index in postion 2 is invails. Array indices must be postive integers or logical values.
I would appricate any help to be able to fix this and get it to run.
Thank You.

请先登录，再回答此问题。

采纳的回答

Torsten
Torsten 2023-7-13
移动：Torsten 2023-7-13
Remove the semicolon behind the line
pivotColumns = rrefA(:,1:end-1);
execute the code and see if it makes sense to insert the displayed matrix in
basisColumnSpace = A(:,pivotColumns);
  7 个评论
显示 5更早的评论
Taylor
Taylor 2023-7-13
I am supposed to be finding the basis for column space.
Torsten
Torsten 2023-7-13
Ok, that's what "licols" does. The maximum number of independent columns of A is a basis for the column space of A.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by