Create a loop for this matrix

1 -1 0 0 0 ...

-1 2 -1 0 0 ...

0 -1 2 -1 0 ...

0 0 -1 2 -1 ...

0 0 0 -1 2 ...

As you can see, this can go on forever, I and would like to find a way to write loop for a dynamic matrix where this just keeps going unitl a certain number 'N'

[dpb]

What's the RH end condition definition?

[Rayden Plasma]

What do you mean?

[David Fletcher]

I think they are asking what the end of the final row should be - I would guess [zeros(N-2,1) -1 1]

[Rayden Plasma]

correct

Answer 1

[David Fletcher]

Just writing something with a loop on the fly (and I stress this is neither robust, nor particularly efficient), but it may be something you can work on in the absence of better answers

clear;
N=10;
base=[1 -1 0;-1 2 -1;0 -1 1];
spliceL=0;
spliceR=N-3;
reform(1,:)=[zeros(1,spliceL) base(1,:) zeros(1,spliceR)];

for iter=2:N-1
    reform(iter,:)=[zeros(1,spliceL) base(2,:) zeros(1,spliceR)];
    spliceL=spliceL+1;
    spliceR=spliceR-1;
end

reform(end+1,:)=[zeros(1,spliceL) base(3,end-1:end) zeros(1,spliceR-1)];

reform

reform = 10×10

1 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 0 0 -1 1

[Rayden Plasma]

Thank you very much!!!