Solving linear systems in Matlab

How do I solve for x in u = A(x+b)? A is a matrix, and the other terms are vectors. Isn't it something like x = A\u - b ? I'm not sure about subtracting the b term, however.

[Bruno Luong]

You need to be careful when the system is undedetermined (usually more unknown than equation)

In some sense x = A\u - b is the worse you can pick.

Answer 1

[Torsten]

x = A\(u-A*b)

But it's the same as

x = A\u - b

[Bruno Luong]

Is it?

A=rand(3,4);
b=rand(4,1);
u=rand(3,1);

% 4 different solutions
x1=A\(u-A*b)

x1 = 4×1

-0.2210 0 -0.6049 -0.1026

x2=A\u-b

x2 = 4×1

-0.0231 -0.8875 -1.3122 1.0476

x3=lsqminnorm(A,u)-b

x3 = 4×1

-0.1409 -0.3592 -0.8912 0.3630

x4=lsqminnorm(A,u-A*b)

x4 = 4×1

-0.2409 0.0891 -0.5339 -0.2181

% All give the same "fitness"
A*x1-u

ans = 3×1

-0.9852 -1.3140 -1.3694

A*x2-u

ans = 3×1

-0.9852 -1.3140 -1.3694

A*x3-u

ans = 3×1

-0.9852 -1.3140 -1.3694

A*x4-u

ans = 3×1

-0.9852 -1.3140 -1.3694

norm(x1)

ans = 0.6522

norm(x2)

ans = 1.8993

norm(x3)

ans = 1.0368

norm(x4) % always the smallest

ans = 0.6314

Answer 2

[Steven Lord]

Define a new variable y such that y = x+b.

Using this new variable simplifies your system to u = A*y.

Solve this simplified system for y using the backslash operator, y = A\u.

Substituing into that solution using the definition of y, we know x+b = A\u.

Subtract b from both sides to get x = (A\u)-b.