Important: The linalg package has been deprecated. Use the superseding packages LinearAlgebra[LinearSolve], instead.

- For information on migrating linalg code to the new packages, see examples/LinearAlgebraMigration.

The function linsolve(A, b) finds the vector x which satisfies the matrix equation $A\\x\\=\\b$. If A has n rows and m columns, then $\mathrm{vectdim}\left(b\right)$ must be n and $\mathrm{vectdim}\left(x\right)$ will be m, if a solution exists.

If $Ax=b$ has no solution or if Maple cannot find a solution, then the null sequence NULL is returned. If $Ax=b$ has many solutions, then the result will use global names (see below) to describe the family of solutions parametrically.

The call linsolve(A, B) finds the matrix X which solves the matrix equation $AX=B$ where each column of X satisfies $\mathrm{Acol}\left(X\,i\right)=\mathrm{col}\left(B\,i\right)$ . If $\mathrm{AX}=B$ has does not have a unique solution, then NULL is returned.

The optional third argument is a name which will be assigned the rank of A.

The optional fourth argument allows you to specify the seed for the global names used as parameters in a parametric solution.  If there is no fourth argument, the default, then the global names _t[1], _t[2], _t[3], ... will be used in the vector case, _t[1][1], _t[1][2], _t[2][1], ... in the matrix case (where _t[1][i] is used for the first column, _t[2][i] for the second, etc).  This is particularly useful when programming with linsolve.  If you declare v as a local variable and then call linsolve with fourth argument v, the resulting parameters (v[1], v[2], ...) will be local to the procedure.

An inert linear solver, Linsolve, is known to the mod function and can be used to solve systems of linear equations (matrix equations) modulo an integer m.

The command with(linalg,linsolve) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{linalg}\right)\:$

$A≔\mathrm{matrix}\left(\left[\left[1\,2\right]\,\left[1\,3\right]\right]\right)\:$

$b≔\mathrm{vector}\left(\left[1\,-2\right]\right)\:$

$\mathrm{linsolve}\left(A\,b\right)$

$\left[\begin{array}{cc}7& \mathrm{-3}\end{array}\right]$

$B≔\mathrm{matrix}\left(\left[\left[1\,1\right]\,\left[-2\,1\right]\right]\right)\:$

$\mathrm{linsolve}\left(A\,B\right)$

$\left[\begin{array}{cc}7& 1\\ \mathrm{-3}& 0\end{array}\right]$

$A≔\mathrm{matrix}\left(\left[\left[5\,7\right]\,\left[0\,0\right]\right]\right)\:$

$b≔\mathrm{vector}\left(\left[3\,0\right]\right)\:$

$\mathrm{linsolve}\left(A\,b\,'r'\right)$

$\left[\begin{array}{cc}\frac{3}{5}-\frac{7{\mathrm{\_t}}_1}{5}& {\mathrm{\_t}}_1\end{array}\right]$

$\mathrm{linsolve}\left(A\,b\,'r'\,v\right)$

$\left[\begin{array}{cc}\frac{3}{5}-\frac{7v_1}{5}& v_1\end{array}\right]$

$A≔\mathrm{matrix}\left(\left[\left[5\,7\right]\,\left[10\,14\right]\right]\right)$

$A≔\left[\begin{array}{cc}5& 7\\ 10& 14\end{array}\right]$

$B≔\mathrm{matrix}\left(\left[\left[3\,0\right]\,\left[6\,0\right]\right]\right)$

$B≔\left[\begin{array}{cc}3& 0\\ 6& 0\end{array}\right]$

$\mathrm{linsolve}\left(A\,B\right)$

$\left[\begin{array}{cc}\frac{3}{5}-\frac{7{\left({\mathrm{\_t}}_1\right)}_1}{5}& -\frac{7{\left({\mathrm{\_t}}_2\right)}_1}{5}\\ {\left({\mathrm{\_t}}_1\right)}_1& {\left({\mathrm{\_t}}_2\right)}_1\end{array}\right]$