How best and most efficient way to build this matrix

2 次查看(过去 30 天)
Hi guys,
I am wondering what is the best and most efficient way to build this matrix using loops and functions? Rather than inputting the numbers manually?
A = [-2 1 0 0 0 0 0 0 0 0
6 -4 1 0 0 0 0 0 0 0
-4 6 -4 1 0 0 0 0 0 0
1 -4 6 -4 1 0 0 0 0 0
0 1 -4 6 -4 1 0 0 0 0
0 0 1 -4 6 -4 1 0 0 0
0 0 0 1 -4 6 -4 0 0 0
0 0 0 0 1 -4 6 -4 1 0
0 0 0 0 0 1 -4 6 -4 1
0 0 0 0 0 0 0 0 1 -2]
A = 10×10
-2 1 0 0 0 0 0 0 0 0 6 -4 1 0 0 0 0 0 0 0 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 0 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 0 0 1 -2
Many thanks
  3 个评论
Matt J
Matt J 2023-9-3
Has A(7,8) deliberately been made 0 instead of 1? If so, it is not clear what the pattern is supposed to be.

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采纳的回答

Matt J
Matt J 2023-9-3
[C,R]=deal(zeros(1,10));
C(1:4)=[-4 6 -4 1];
R(1:2)=[-4,1];
A=toeplitz(C,R);
A(1)=-2;
A(end,:)=flip(A(1,:))
A = 10×10
-2 1 0 0 0 0 0 0 0 0 6 -4 1 0 0 0 0 0 0 0 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 1 -4 6 -4 1 0 0 0 0 0 0 0 0 1 -2

更多回答(1 个)

Torsten
Torsten 2023-9-3
移动:Torsten 2023-9-3
Experiment with "spdiags":
If first and last row appear different from your matrix, you can change them subsequently.

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