If all the variables of a matrix each have their own leading ones and the last row is all zeros is it still a unique solution?

Question

The original augmented matrix is $$ \begin{pmatrix} \begin{array}{ccc|c} 3 & 3 & 6 & 6 \\ 2 & 2 & 2 & 0 \\ -3 & -3 & -5 & -4 \\ -2 & -1 & -1 & 2 \\ \end{array} \end{pmatrix} $$ and the reduced row echelon form is \begin{pmatrix} \begin{array}{ccc|c} 1 & 0 & 0 & -6 \\ 0 & 1 & 0 & 12 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ \end{array} \end{pmatrix}

Is the unique solution \begin{equation} \left (\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \end{array}\right ) \ = \left( \begin{array}{c} -6 \\ 12 \\ -2 \\ \end{array} \right) \end{equation} or is it an infinite number of solutions?

Answer 1

Yes. If you have more equations than you do unknowns, than any subsequent rows in the matrix should be a zero row, since you can always reduce them using the leading ones from the above rows. If there were more than one zero rows in your RREF, then you would have infinitely many solutions, since then $x_3$ would be free.

Answer 2

It indeed has a unique solution; a non-homogeneous linear system has a solution if and only if the matrix of the l.h.s. and the augmented matrix have the same rank. This common rank is the codimension of the affine subspace of solutions.

Here the common rank is the maximal possible rank – $3$, so the subspace of solutions has codimension $3$. In a $3$-dimensional space, this means it has dimension $0$, i.e. it is a single point.