Interpret a matrix in reduced row echelon form

mark

Suppose that
$$
\left(\begin{matrix}
0 & 1 & 0 & 0 & 2 & 3 & 0 & 2 \\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 3 \\
0 & 0 & 0 & 1 & 1 & 2 & 0 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1
\end{matrix}\right)
$$
is the reduced row echelon form of an augmented matrix representing a linear system of equations. Write down the solution or solutions of that system.

Student39

The there are infinitely many solutions to the the matrix as follows:

\begin{align}x_{1} & = x_{1}\\
x_{2} &= -2x_{5} - 3x_{6} + 2\\
x_{3} &= 3-x_{5}\\
x_{4} &=- x_{5} - 2x_{6} + 4\\
x_{5} &= 3 - x_{3}\\
x_{6} &= 4 - x_{5} - x_{4}\\
x_{7} &= 1\\
\end{align}

mark

@Student39 We typically take the dependent variables to correspond to the pivots - i.e. the leading ones.

Student39

So the following because $x_5$ & $x_6$ are independent:

\begin{align}
x_{2} &= -2x_{5} - 3x_{6} + 2\\
x_{3} &= 3-x_{5}\\
x_{4} &= - x_{5} - 2x_{6} + 4\\
x_{7} &= 1\\
\end{align}

mark

@Student39 said:
So the following because $x_5$ & $x_6$ are independent:

\begin{align}
x_{2} &= -2x_{5} - 3x_{6} + 2\\
x_{3} &= 3-x_{5}\\
x_{4} &= - x_{5} - 2x_{6} + 4\\
x_{7} &= 1\\
\end{align}

That looks much better! Note that tautological statements $x_1=x_1$ are not necessary and I edited accordingly.