I want to make an identity matrix

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Qonitat
Qonitat il 5 Mar 2023
Modificato: John D'Errico il 5 Mar 2023
i want to make a matrix that look like this
How could i achieve this efficiently .Thanks in advace

Risposte (2)

the cyclist
the cyclist il 5 Mar 2023
I've made some assumptions about the pattern, but I expect it is what you want.
This should be highly efficient. The algorithm is a bit obfuscated.
% Input
N = 4;
% Start with all 0s
M = zeros(N,N*(N+1));
% Fill in the 1s
M(1:N*N+1:end-N*N) = 1;
% Fill in the -1s
M(:,N*N+1) = -1;
% Display the result
disp(M)
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 -1 0 0 0

John D'Errico
John D'Errico il 5 Mar 2023
Modificato: John D'Errico il 5 Mar 2023
These are starting to look like homework questions.
Did you not read the answer to the last question you asked? You could not use that idea to build this matrix?
A = kron([eye(4),zeros(4,1)],[1 0 0 0])
A = 4×20
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
That gets you almost there, with the first 16 columns correct, and a total of 20 columns. Then it would be simple enough to stuff in the -1 elements in column 17.
A(:,17) = -1
A = 4×20
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 -1 0 0 0
spy(A)

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