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See the recommended documentation of this function

Scilab help >> Scilab > Scilab keywords > backslash (\)

backslash (\)

left matrix division.

x=A\b

Description

Backslash denotes left matrix division. x=A\b is a solution to A*x=b.

If A is square and nonsingular 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 (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 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
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

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());

• slash — (/) right division and feed back
• inv — matrix inverse
• pinv — pseudoinverse
• percent — (%) special character
• ieee — set floating point exception mode
• linsolve — linear equation solver
• umfpack — solve sparse linear system