What would the solution be if I cannot simplify my solution to find more basic variables(pivots)? 4x3 matrix $$x - 5y + 4z = -3$$
$$2x - 7y + 3z = -2$$
$$-2x + y +  7z = -1$$
The furthest I was able to reach before coming to the conclusion that I could not move any further was:
Row 1 => $x - 5y + 4z = -3$
Row 2 => $9y + 15z = -7$
Row 3 => $8y + 5z = 1$
Please explain to me how I would be able to express my solution in the simplest form! Also let me know what type of solution this system would contain (independent, dependent, inconsistent)
 A: Here is the reduction of the augmented matrix.
Column 1
$$
  \left[
\begin{array}{rcc}
 1 & 0 & 0 \\
 -2 & 1 & 0 \\
 2 & 0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{rrc|ccc}
 1 & -5 & 4 & 1 & 0 & 0 \\
 2 & -7 & 3 & 0 & 1 & 0 \\
 -2 & 1 & 7 & 0 & 0 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{crr|rcc}
 \boxed{1} & -5 & 4 & 1 & 0 & 0 \\
 0 & 3 & -5 & -2 & 1 & 0 \\
 0 & -9 & 15 & 2 & 0 & 1 \\
\end{array}
\right]
$$
Column 2
$$
  \left[
\begin{array}{ccc}
 1 & \frac{5}{3} & 0 \\
 0 & \frac{1}{3} & 0 \\
 0 & 3 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{crr|rcc}
 \boxed{1} & -5 & 4 & 1 & 0 & 0 \\
 0 & 3 & -5 & -2 & 1 & 0 \\
 0 & -9 & 15 & 2 & 0 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{crr|rcc}
 \boxed{1} & 0 & -\frac{13}{3} & -\frac{7}{3} & \frac{5}{3} & 0 \\
 0 & \boxed{1} & -\frac{5}{3} & -\frac{2}{3} & \frac{1}{3} & 0 \\
 0 & 0 & 0 & -4 & 3 & 1 \\
\end{array}
\right]
$$
Least squares solution
The least squares solution to $$\mathbf{A}x = b$$ is 
$$
 x = \frac{1}{5827}
%
 \left[
\begin{array}{r}
 -257 \\ 1369 \\ -1168 \\
\end{array}
\right]
$$
The least squares residual errors is
$$
\begin{align}
  \mathbf{A} x - b &= \mathbf{0} \\
%
 \left[
\begin{array}{rrc}
 1 & -5 & 4 \\
 2 & -7 & 3 \\
 -2 & 1 & 7 \\
\end{array}
\right]
\frac{1}{5827}
%
 \left[
\begin{array}{r}
 -257 \\ 1369 \\ -1168 \\
\end{array}
\right]
%
 - 
 \left[
\begin{array}{r}
 -3 \\ -2 \\ -1 \\
\end{array}
\right]
%
&=
\frac{1}{26}
 \left[
\begin{array}{r}
 20 \\ -15 \\ -5 \\
\end{array}
\right]
%
\end{align}
$$
