Least square magnitude solution

is this a right approach to calculate the least square magnitude x* to this problem? Any help will be greatly appreciated. Thanks

A = [1 2 3 4 5; 6 7 8 9 0];

 b = [1;2];

x = lsqr(A,b)

e2 = b - A*x

norm(e2)

Answer 1

[Torsten]

A = [1 2 3 4 5; 6 7 8 9 0];
b = [1;2];
x = lsqminnorm(A,b).'

x = 1×5

0.0400 0.0560 0.0720 0.0880 0.0560

norm(x)

ans = 0.1442

norm(A*x.'-b)

ans = 0

if you search for the solution for which x has minimum norm.

If you only search for an arbitrary least-squares solution (i.e. for an x for which norm(e2) is minimum), simply use

x = (A\b).'

x = 1×5

0 0 0 0.2222 0.0222

norm(x)

ans = 0.2233

norm(A*x.'-b)

ans = 2.2204e-16

[Cesar Cardenas]

Right thank so much