Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-4.

a=[1 1 3;2 0 2;1 1 0;];
b=[5;6;7]
ab=[a b]
 %pivot 1,1
 %
 if ab(1,1)< abs(max(ab(:,1)))
     piv=ab(1,:)
     ab(1,:)=ab(2,:)
     ab(2,:)=piv
 
   
  ab(2,1)=ab(2,:)-ab(2,1)/ab(1,1)*ab(1,:)
  ab(3,1)=ab(3,:)-ab(2,1)/ab(1,1)*ab(1,:)
 end

Answer 1

[Walter Roberson]

ab(2,1)=ab(2,:)-ab(2,1)/ab(1,1)*ab(1,:)

ab(2,1) is a scalar and ab(1,1) is a scalar, so ab(2,1)/ab(1,1) is a scalar. So at the end of the expression, you are multiplying the vector ab(1,:) by a scalar, getting back a vector.

ab(2,:) is a vector.

You are subtracting a vector from a vector, which is an operation that returns a vector.

So the right hand side is a vector.

The destination ab(2,1) is, however, a scalar location. You cannot store a vector into a scalar location.

[Rabia Sonmez]

yeah I solved it. Thank you:)