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Growing eye matrix as per the size of eye

1 Ansicht (letzte 30 Tage)
Babu Thomas
Babu Thomas am 2 Jun. 2022
Kommentiert: Babu Thomas am 2 Jun. 2022
I have an eye matrix of size eye(3). I need to increase the size of matrix as below so that eye matrix has to repeat 2 times then 3 times. Can anyone help

Akzeptierte Antwort

Chunru
Chunru am 2 Jun. 2022
a = tril(repmat(eye(3), [3, 3]))
a = 9×9
1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1

Weitere Antworten (3)

DGM
DGM am 2 Jun. 2022
There are a number of ways, but this is what I'd do:
A = toeplitz([1 0 0 1 0 0 1 0 0],[1 0 0 0 0 0 0 0 0])
A = 9×9
1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1

Bruno Luong
Bruno Luong am 2 Jun. 2022
Bearbeitet: Bruno Luong am 2 Jun. 2022
tril(repmat(eye(3),3))
ans = 9×9
1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1

KSSV
KSSV am 2 Jun. 2022
REad about diag
I = diag(repelem(1,1,9))+diag(repelem(1,1,6),-3)+diag(repelem(1,1,3),-6)
I = 9×9
1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1

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