# How to add, multiply and subtract two matrices have different length ?

14 views (last 30 days)
omar th on 13 Mar 2022
Answered: Image Analyst on 26 Mar 2022
If I have matrix A =[2 2 1; 1 2 5; 1 2 3], B=[1 2; 1 1],
How can I add or subtract these matrices while they have different length ?
##### 1 CommentShowHide None
Torsten on 13 Mar 2022
Mathematical, there are no such operations for matrices of different sizes.
So you first will have to explain us how you intend to define these "additions and subtractions".

Star Strider on 13 Mar 2022
While adding two matrices of different dimensions is not defined mathematically, it is possible to add or subtract them by indexing into them, however it is first necessary to define what elements are to be affected in the larger matrix.
Here is one approach —
A = [2 2 1; 1 2 5; 1 2 3]
A = 3×3
2 2 1 1 2 5 1 2 3
B = [1 2; 1 1]
B = 2×2
1 2 1 1
idx = 1:numel(B); % Using The Most Obvious Index Vector ...
C = A;
C(idx) = A(idx) + B(idx)
C = 3×3
3 3 1 2 2 5 3 2 3
idx = randperm(numel(A), numel(B)) % Using A Different, Random, Index Vector ...
idx = 1×4
4 6 5 1
C = A;
C(idx) = A(idx) + B(1:numel(B))
C = 3×3
3 3 1 1 4 5 1 3 3
So in a limited sense it is possible, however I have no way of knowing if either of these produces the desired result.
.
Image Analyst on 26 Mar 2022
@omar th you forgot to read Community Guidelines or these posting guidelines:
So, because of that you forgot to state what the desired results are! I guess you also missed Star's gentle hint. So, what are they? What elements get subtracted from what elements, and what elements are ignored during the operation?

### More Answers (1)

Image Analyst on 26 Mar 2022
Perhaps this is what you want -- to subtract the upper left parts that overlap.
A = [2 2 1; 1 2 5; 1 2 3]
B = [1 2; 1 1]
[rowsA, colsA] = size(A)
[rowsB, colsB] = size(B)
maxRow = max([rowsA, rowsB])
maxCol = max([colsA, colsB])
% Pad dimensions if nexessary
if rowsA > rowsB
B(maxRow, end) = 0;
[rowsB, colsB] = size(B) % Update size
elseif rowsB > rowsA
A(maxRow, end) = 0;
[rowsA, colsA] = size(A) % Update size
end
if colsA > colsB
B(end, maxCol) = 0;
[rowsB, colsB] = size(B) % Update size
elseif colsB > colsA
A(end, maxCol) = 0;
[rowsA, colsA] = size(A) % Update size
end
% Let's see what they are now:
A
B
% Now do the subtraction of the upper left overlapping parts.
C = A - B
You get:
A =
2 2 1
1 2 5
1 2 3
B =
1 2 0
1 1 0
0 0 0
C =
1 0 1
0 1 5
1 2 3