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See the recommended documentation of this function
backslash
(\) left matrix division.
Calling Sequence
x=A\b
Description
Backslash denotes left matrix division.
x=A\b
is a solution to A*x=b
.
If A
is square and non-singular x=A\b
(uniquely defined) is equivalent to x=inv(A)*b
(but the computations are much cheaper).
If A
is not square, x
is a least square solution,
i.e. norm(A*x-b)
is minimal (Euclidean norm). If A
is full
column rank, the least square solution, x=A\b
, is uniquely
defined (there is a unique x
which minimizes norm(A*x-b)
).
If 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 x=pinv(A)*b
).
A.\B
is the matrix with (i,j)
entry A(i,j)\B(i,j)
.
If A
(or B
) is a scalar A.\B
is equivalent to
A*ones(B).\B
(or A.\(B*ones(A))
.
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 /
.
Examples
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 // if rank is deficient 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 = zeros(size(A,1),1); tic(); res = umfpack(A,'\',b); mprintf('\ntime needed to solve the system with umfpack: %.3f\n',toc()); tic(); res = linsolve(A,b); mprintf('\ntime needed to solve the system with linsolve: %.3f\n',toc()); tic(); res = A\b; mprintf('\ntime needed to solve the system with the backslash operator: %.3f\n',toc());
See Also
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