The LinearSolve command returns the Vector x that satisfies the linear system $A·x=b$.

If b is omitted, then the linear system is interpreted from the first argument, which is taken to be the augmented Matrix in which the last column provides the right-hand side data.

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{LinearAlgebra}\right]\right)\:$

$A≔⟨⟨1\,2\,3⟩\left|⟨4\,5\,6⟩\right|⟨7\,8\,0⟩⟩\:$

$b≔⟨3\,2\,-2⟩\:$

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

$\left[\begin{array}{c}\mathrm{-2}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$

$M≔⟨⟨1\,1\,1\,4⟩\left|⟨1\,1\,-2\,1⟩\right|⟨3\,1\,1\,8⟩\left|⟨-1\,1\,-1\,-1⟩\right|⟨0\,1\,1\,0⟩⟩\:$

$\mathrm{M\_sol}≔\mathrm{LinearSolve}\left(M\right)$

$\mathrm{M\_sol}≔\left[\begin{array}{c}\frac{25}{6}\\ \frac{4}{3}\\ -\frac{5}{2}\\ \mathrm{-2}\end{array}\right]$

$M\left[1..-1,1..-2\right]·\mathrm{M\_sol}=M\left[1..-1,-1\right]$

$\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right]$

$R≔⟨⟨1\,2\,3⟩\left|⟨4\,5\,7⟩\right|⟨7\,8\,9⟩|⟨10\,11\,12⟩⟩\:$

$z≔⟨3\,2\,0⟩\:$

$\mathrm{LinearSolve}\left(R\,z\right)$

$\left[\begin{array}{c}-\frac{7}{6}+\frac{{\mathrm{\_t1}}_4}{2}\\ \mathrm{-1}\\ \frac{7}{6}-\frac{3{\mathrm{\_t1}}_4}{2}\\ {\mathrm{\_t1}}_4\end{array}\right]$