Find and interpret the reduced row echelon form of a matrix

mark

Let
$$M = \left(
\begin{matrix}
2 & -1 & 3 & 2 \\
2 & 1 & 6 & 7 \\
-2 & 1 & 0 & 3
\end{matrix}\right).$$

1. Find the reduced row echelon form of $M$.
2. Assuming that $M$ is an augmented matrix representing a linear system of equations, write down the solution or solutions of that system.

Student17

The reduced row echelon form can be shown as $$M = \left(
\begin{matrix}
1 & 0 & 0 & -3/2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 5/3
\end{matrix}\right).$$

Therefore, the solutions can be shown as: $$\begin{align}
x_{1}&=\dfrac{-3}{2}\\
x_{2}&=0\\
x_{3}&=\dfrac{5}{3}
\end{align}$$

mark

@Student17 Yes - that's correct! It might be worth mentioning that you can show steps using a notation like so:

$$
\left(
\begin{matrix}
2 & -1 & 3 & 2 \\
2 & 1 & 6 & 7 \\
-2 & 1 & 0 & 3
\end{matrix}
\right)
\xrightarrow[-R1+R2]{R1+R3}
\left(
\begin{matrix}
2 & -1 & 3 & 2 \\
0 & 2 & 3 & 5 \\
0 & 0 & 3 & 5
\end{matrix}
\right)
\xrightarrow[R2+R1]{\frac{1}{2}R2}
\left(
\begin{matrix}
2 & 0 & 9/2 & 9/2 \\
0 & 1 & 3/2 & 5/2 \\
0 & 0 & 3 & 5
\end{matrix}
\right)
$$
$$
\xrightarrow[-\frac{1}{2}R3+R2]{-\frac{3}{2}R3+R1}
\left(
\begin{matrix}
2 & 0 & 0 & -3 \\
0 & 1 & 0 & 0 \\
0 & 0 & 3 & 5
\end{matrix}
\right)
\xrightarrow[\frac{1}{2}R1]{\frac{1}{3}R3}
\left(
\begin{matrix}
1 & 0 & 0 & -3/2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 5/3
\end{matrix}
\right)
$$