MATLAB code to create a pattern of matrix

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ANDREW B I
ANDREW B I 2021 年 2 月 24 日
コメント済み: Rik 2021 年 2 月 24 日
Diagonal
for i=1 to 2n+2,
for j=1 to 2n+2,
aij=0 ; when i=j
end
1st row
for j= 2 to n+2
aij=1
for j=n+3 to 2n+2
aij=0
2nd row
for j=3 to n+2
a2j=0
for j=n+3 to 2n+2
a2j=1
3rd row
a34=-1, a3,n+3 = -1
for j=5 to n+2; n+4 to 2n+2
4th to n+1 rows
For i=4 to n+1
For j=1 to 2n+2 and i<j
When j=i+1
aij = -1
else aij=0
end
(n+2)th row
for i = n+2 and i<j
when j=2n+2
aij=-1
else aij= 0
(n+3 to 2n+1) rows
for i=n+3 to 2n+1 and i<j
when j=i+1
aij=1
else aij=0
lower diagonal elements
for i>j
aij=-aij
  2 件のコメント
Rik
Rik 2021 年 2 月 24 日
It seems you already have an answer to your other question.
If you don't, please follow my advice there.

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