'Solve linear systems Ax=b'
'You can use x=A\b to get one solution'
A=[1,1,0;2,3,4]
b=[3;-2]
x=A\b
'OK if there is only one. But we want them all!'
'Example from p. 538 of textbook'
'Either use solve'
[w,x,y,z]=solve(
'3*x+y+7*z+2*w=13',
'2*x-4*y+14*z-w=-10',
'5*x+11*y-7*z+8*w=59',
'2*x+5*y-4*z-3*w=39')
'or express the system as an augmented matrix (better!)'
M6=[3,1,7,2,13;
2,-4,14,-1,-10;
5,11,-7,8,59;
2,5,-4,-3,39]
'and calculate the rref!'
%You can do row operations
'Subtract row 4 from row 2' 
M6(2,:)=M6(2,:) - M6(4,:)
'Subtract row 4 from row 1, then 5 times 1 from 3' 
M6(1,:)=M6(1,:) - M6(4,:); %the ;
% suppresses output
M6(3,:)=M6(3,:) - 5*M6(1,:)
'and so on. But there is a function rref that does all this for you.'
M7=rref(M6)
'We can consider the system solved here'.
%One final note. As M6[i,:] is row i of M6,
%so M6[:,j] is column j.