 Null Space - Maple Help

LinearAlgebra[Generic]

 NullSpace
 compute the nullspace of a Matrix Calling Sequence NullSpace[F](A) Parameters

 F - the domain of computation, a field A - rectangular Matrix over values in F Description

 • NullSpace(A) returns a basis for the linear system A x = 0 over the field F as a set of Vectors B = {b1, b2, ...}.
 • The (indexed) parameter F, which specifies the domain of computation, a field, must be a Maple table/module which has the following values/exports:
 F: a constant for the zero of the ring F
 F: a constant for the (multiplicative) identity of F
 F[+]: a procedure for adding elements of F (nary)
 F[-]: a procedure for negating and subtracting elements of F (unary and binary)
 F[*]: a procedure for multiplying two elements of F (commutative)
 F[/]: a procedure for dividing two elements of F
 F[=]: a boolean procedure for testing if two elements in F are equal Examples

 > $\mathrm{with}\left(\mathrm{LinearAlgebra}\left[\mathrm{Generic}\right]\right):$
 > $Q\left[\mathrm{0}\right],Q\left[\mathrm{1}\right],Q\left[\mathrm{+}\right],Q\left[\mathrm{-}\right],Q\left[\mathrm{*}\right],Q\left[\mathrm{/}\right],Q\left[\mathrm{=}\right]≔0,1,\mathrm{+},\mathrm{-},\mathrm{*},\mathrm{/},\mathrm{=}$
 ${{Q}}_{{0}}{,}{{Q}}_{{1}}{,}{{Q}}_{{\mathrm{+}}}{,}{{Q}}_{{\mathrm{-}}}{,}{{Q}}_{{\mathrm{*}}}{,}{{Q}}_{{\mathrm{/}}}{,}{{Q}}_{{\mathrm{=}}}{≔}{0}{,}{1}{,}{\mathrm{+}}{,}{\mathrm{-}}{,}{\mathrm{*}}{,}{\mathrm{/}}{,}{\mathrm{=}}$ (1)
 > $A≔\mathrm{Matrix}\left(\left[\left[1,2,3\right],\left[1,3,5\right],\left[0,1,2\right]\right]\right)$
 ${A}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {1}& {3}& {5}\\ {0}& {1}& {2}\end{array}\right]$ (2)
 > $\mathrm{NullSpace}\left[Q\right]\left(A\right)$
 $\left\{\left[\begin{array}{c}{1}\\ {-2}\\ {1}\end{array}\right]\right\}$ (3)
 > $A≔\mathrm{Matrix}\left(\left[\left[1,2,3\right],\left[2,4,6\right],\left[-1,-2,-3\right]\right]\right)$
 ${A}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {2}& {4}& {6}\\ {-1}& {-2}& {-3}\end{array}\right]$ (4)
 > $\mathrm{NullSpace}\left[Q\right]\left(A\right)$
 $\left\{\left[\begin{array}{c}{-2}\\ {1}\\ {0}\end{array}\right]{,}\left[\begin{array}{c}{-3}\\ {0}\\ {1}\end{array}\right]\right\}$ (5)