Switching rows in matrix

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Rachel McMurphy
Rachel McMurphy 2019-12-3
编辑： cater re 2019-12-4
First off, here is the code I have:
function m = move(M,i,j)
[x, y] = size(M); %n is rows and m is columns
submatrix = M([i:x],[1:y]); %creates submatrix of ith row down
column_j = submatrix(:,j); %gives jth column
row = find(column_j,1); %gives row number of first nonzero
i = M(i,:); %gives ith row of M
k = M(row+1,:); %gives row that has nonzero in M
M([i k],:) = M([k i],:)
end
For reference, all the little steps are required to show in my assignment and I've done my best to explain what their outputs are. For matrix
M = [1 3 2 -4 1 10 -3;0 0 0 -2 0 4 0;0 0 0 -4 1 7 -2;0 0 -3 6 -2 -10 4];
I have to find the submatrix of M from ith row down and the subcolumn of choice (lines 3 and 4), find the row number (line 5), and then take that specific row (which I got in line 7) and switch it with the ith row. So by doing:
move(M,2,3)
with the above matrix, the output should be
M =[1 3 2 -4 1 10 -3;0 0 -3 6 -2 -10 4;0 0 0 -4 1 7 -2;0 0 0 -2 0 4 0]
but my last step (line 8) which should switch the rows is bringing up an error. If anything is confusing, please let me know, I can try to explain the assignment as best as I can

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回答（1 个）

cater re
cater re 2019-12-4
编辑：cater re 2019-12-4
Error is erased, but you need to fix this code. The code don't make result to your intend.
M = [1 3 2 -4 1 10 -3;0 0 0 -2 0 4 0;0 0 0 -4 1 7 -2;0 0 -3 6 -2 -10 4];
i=2
j=3
[x, y] = size(M); %n is rows and m is columns
submatrix = M([i:x],[1:y]); %creates submatrix of ith row down
column_j = submatrix(:,j); %gives jth column
row = find(column_j,1); %gives row number of first nonzero
l = M(i,:); %gives ith row of M
k = M(row+1,:); %gives row that has nonzero in M
M(i,:) = k
M(j,:) = l
switching:
M = [1 3 2 -4 1 10 -3; 0 0 0 -2 0 4 0 ; 0 0 0 -4 1 7 -2 ; 0 0 -3 6 -2 -10 4];
i=2
j=4
temp = M(j,:)
M(j,:) = M(i,:)
M(i,:) = temp

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