Verify my For Loops

Hey guys! I am still kind of getting a hang of Matlab and I have a quick favor to ask you guys. The question is to create a double for loop for the following 3 × 4 matrix with ijth matrix element: A(i,j)=i+j, 1≤i≤3, 1≤j≤4 Construct this matrix in MATLAB using the following four methods: a. Using a double for-loop. b. Using a for-loop that builds the matrix row by row. c. Using a for-loop that builds the matrix column by column.

I have already done this but I am getting sort of an odd answer, would someone please verify my work?

i=3;
j=4;
A=zeros(i,j);
for k=1:i, 
    for z=1:j
    A(k,z)=i+j;
    end
end
A
%Part B
for k=1:i
    A(k)=i+j;
end
A
%Part C
for z=1:j
    A(z)=i+j;
end
A

oh and this are the arrays I got:

A =

     7     7     7     7
     7     7     7     7
     7     7     7     7

A =

     7     7     7     7
     7     7     7     7
     7     7     7     7

A =

     7     7     7     7
     7     7     7     7
     7     7     7     7

Answer 1

[James Tursa]

The reason you are getting all 7's in your answer is because i+j is always 7, so you are assigning 7 to each element. And the indexing is not correct for your second and third parts.

For the double for loop, you need to make the (k,z) element depend on the variables k and z which are used for the loop indexes, not i and j which are not used for the loop indexes. E.g.,

    A(k,z) = k + z;

Had you used i and j for the loop indexes, then you could have used i + j in the above line.

For the building by row part, you need the assignment to be this for the k'th row:

    A(k,:) = SOMETHING;  % <-- You need to figure out how to fill in the SOMETHING

For the building by column part, you need the assignment to be this for the z'th column:

    A(:,z) = SOMETHING;  % <-- You need to figure out how to fill in the SOMETHING

So, I have fixed the first part for you, and partially fixed the second and third parts. You just need to figure out how to make a row of the appropriated numbers for the second part, and how to make a column of the appropriate numbers for the third part.

[scarlet knight]

Thank you so much, yeah I figured something was way off. I'm glad you were kind enough to explain what I did wrong without making me seem like a dummy haha.

[scarlet knight]

Okay so this is what I ended up doing because I could not figure out a function for A(k,:)

k=1;
A=zeros(k);
for k=1:i
  A(k,1)=k+1;
  A(k,2)=k+2;
  A(k,3)=k+3;
  A(k,4)=k+4;
end
A

[James Tursa]

Think of it this way:

A(k,:) is the k'th row, and it has "j" number of elements. The elements are as follows:

k+1 , k+2 , k+3 , ... , k+j

So how to assign those row elements all at once? In MATLAB m-code a row vector with these elements can be built as follows:

k + (1:j)

The (1:j) part will create a row vector with the elements 1,2,...,j and the k+ part will add k to each element of this row vector. So the row assignment becomes, in one line,

A(k,:) = k + (1:j);

For the building by z'th column part you will need to do something similar but constructing a column vector with the elements 1+z,2+z,...,i+z instead of a row vector. Hint: Since (1:i) is a row vector, its transpose (1:i)' is a column vector.

Answer 2

[John BG]

N=3
M=4
A=zeros(N,M)
for i=1:1:N 
    for j=1:1:M
    A(i,j)=i+j
    end
end

N=3
M=4
A=zeros(N,M)
j=1
while j<M+1
for i=1:1:N 
     A(i,j)=i+j
 end
   j=j+1
end

N=3
M=4
A=zeros(N,M)
i=1
while i<N+1
for j=1:1:M 
     A(i,j)=i+j
 end
   i=i+1
end

And the MATLAB way would be:

N=3;M=4;A=zeros(N,M)
[I,J]=ind2sub(size(A),find(A==0))
A=I+J
reshape(A,N,M)