Indexing in a matrix vs vector

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Luisa
Luisa on 12 Nov 2025 at 12:15
Commented: dpb on 12 Nov 2025 at 16:43
Why can we write A(1,2) for a nx2 matrix, with n>1, and we cannot write i=1; x(i,2) for a vector x = [x1 x2]?
It only accepts x(i+1), but I need to iterate the index i over the rows and not the columns, for any nx2 matrix with n>=1.
Thank you in advance.

Accepted Answer

dpb
dpb on 12 Nov 2025 at 12:35
There's nothing preventing addressing a row vector by its row, column indices...
x=1:3;
for i1=1:height(x)
for j1=1:width(x)
disp([i1 j1 x(i1,j1)])
end
end
1 1 1 1 2 2 1 3 3
One only gets in trouble when trying to go outside array bounds...
x(2,1)
Index in position 1 exceeds array bounds. Index must not exceed 1.
  4 Comments
Luisa
Luisa on 12 Nov 2025 at 16:21
Edited: Luisa on 12 Nov 2025 at 16:30
Yes, I had made a syntax error in the code in another place, and it contested as "Index in position 1 exceeds array bounds. Index must not exceed 1."
Thank you so much.
dpb
dpb on 12 Nov 2025 at 16:43
Glad to help and that you found it...enjoy!

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More Answers (1)

Image Analyst
Image Analyst on 12 Nov 2025 at 13:30
i = 1;
x = [10, 20]
x = 1×2
10 20
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x(i, 2) % Display second column of first (and only) row of a row vector.
ans = 20
I'm not seeing the problem. It works. You must be describing something different than your actual code.
Contrary to what you said, it will not accept x(i+1, 2), which is x(2,2), because there is no second row of x. Remember row is the first index and column is the second index.
I encourage you to use descriptive names like row and col instead of i and j when using indexes in arrays. It will avoid a lot of mistakes.
  1 Comment
Luisa
Luisa on 12 Nov 2025 at 16:24
Thank you, I had made a syntax error in the code in another place.

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