Solve the two systems of linear equations below.
\[ \begin{array}{rcrcrcr} -x & - & y & + & 3z & = & 1 \\ -2x & + & y & - & 3z & = & 4 \\ -x & - & 2y & + & 2z & = & -1 \end{array}\quad\text{and}\quad \begin{array}{rcrcrcr} -x & - & y & + & 3z & = & 24 \\ -2x & + & y & - & 3z & = & -8 \\ -x & - & 2y & + & 2z & = & -14 \end{array} \]First of all we observe that both systems of linear equations have the same coefficient matrix
\[ A\ =\ \left[\begin{array}{rrr} -1 & -1 & 3 \\ -2 & 1 & -3 \\ -1 & -2 & 2 \end{array} \right] \]This matrix is square-shaped and of size \((3,3)\). If it happens to be invertible, we can use its inverse to conveniently solve both systems. Indeed, we find
\[ A^{-1}\ =\ \dfrac{1}{12}\, \left[\begin{array}{rrr} -4 & -4 & 0 \\ 7 & 1 & -9 \\ 5 & -1 & -3 \end{array}\right] \]Consequently, the first system has the unique solution
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \dfrac{1}{12}\, \left[\begin{array}{rrr} -4 & -4 & 0 \\ 7 & 1 & -9 \\ 5 & -1 & -3 \end{array}\right] \left[\begin{array}{r} 1 \\ 4 \\ -1 \end{array}\right]\ =\ \dfrac{1}{3}\, \left[\begin{array}{r} -5 \\ 5 \\ 1 \end{array}\right] \]While, the second system has the unique solution
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \dfrac{1}{12}\, \left[\begin{array}{rrr} -4 & -4 & 0 \\ 7 & 1 & -9 \\ 5 & -1 & -3 \end{array}\right] \left[\begin{array}{r} 24 \\ -8 \\ -14 \end{array}\right]\ =\ \dfrac{1}{6}\, \left[\begin{array}{r} -32 \\ 143 \\ 85 \end{array}\right] \]