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left matrix division.
Backslash denotes left matrix division.
x=A\b is a solution to
A is square and nonsingular
x=A\b (uniquely defined) is equivalent to
x=inv(A)*b (but the computations are much cheaper).
A is not square,
x is a least square solution.
norm(A*x-b) is minimal (euclidian norm). If
A is full
column rank, the least square solution,
x=A\b, is uniquely
defined (there is a unique
x which minimizes
A is not full column rank, then the least square
solution is not unique, and
x=A\b, in general, is not the solution
with minimum norm (the minimum norm solution is
A.\B is the matrix with
B) is a scalar
A.\B is equivalent to
A\.B is an operator with no predefined meaning. It may be used
to define a new operator (see overloading) with the same precedence as * or /.
A=rand(3,2);b=[1;1;1]; x=A\b; y=pinv(A)*b; x-y A=rand(2,3);b=[1;1]; x=A\b; y=pinv(A)*b; x-y, A*x-b, A*y-b A=rand(3,1)*rand(1,2); b=[1;1;1]; x=A\b; y=pinv(A)*b; A*x-b, A*y-b A=rand(2,1)*rand(1,3); b=[1;1]; x=A\b; y=pinv(A)*b; A*x-b, A*y-b // A benchmark of several linear solvers [A,descr,ref,mtype] = ReadHBSparse(SCI+"/modules/umfpack/examples/bcsstk24.rsa"); b = 0*ones(size(A,1),1); tic(); res = umfpack(A,'\',b); printf('\ntime needed to solve the system with umfpack: %.3f\n',toc()); tic(); res = linsolve(A,b); printf('\ntime needed to solve the system with linsolve: %.3f\n',toc()); tic(); res = A\b; printf('\ntime needed to solve the system with the backslash operator: %.3f\n',toc());
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