Linear Algebra

Determine Whether Matrices are in Reduced Row Echelon Form, and Find Solutions of Systems

The possibilities for the solution set of a system of linear equations

Problem 648

Determine whether the following augmented matrices are in reduced row echelon form, and calculate the solution sets of their associated systems of linear equations.

(a) $\left[\begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 6 \end{array} \right]$.

(b) $\left[\begin{array}{rrr|r} 1 & 0 & 3 & -4 \\ 0 & 1 & 2 & 0 \end{array} \right]$.

(c) $\left[\begin{array}{rr|r} 1 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right]$.

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Solution.

(a) $\left[\begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 6 \end{array} \right]$

The first matrix is in reduced row echelon form.

For this matrix, we can read off the solution $x_1 = 2, x_2 = -3, x_3 = 6$.

(b) $\left[\begin{array}{rrr|r} 1 & 0 & 3 & -4 \\ 0 & 1 & 2 & 0 \end{array} \right]$

The second matrix is in reduced row echelon form.

For this matrix, the variable $x_3$ is a free variable because there are no leading 1s in the 3rd column.

The solution set can be expressed as $x_1 = -4 – 3 x_3$, $x_2 = -2 x_3$, and $x_3$ can be any real number.

(c) $\left[\begin{array}{rr|r} 1 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right]$

The third matrix is not in reduced echelon form because the bottom-left entry is $1$, not $0$, so we first use elementary row operations to put it in this form.
\begin{align*}
\left[\begin{array}{rr|r} 1 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right] \xrightarrow{R_2 – R_1} \left[\begin{array}{rr|r} 1 & 2 & 0 \\ 0 & -1 & -1 \end{array} \right]\\[6pt] \xrightarrow{(-1) R_2 } \left[\begin{array}{rr|r} 1 & 2 & 0 \\ 0 & 1 & 1 \end{array} \right] \xrightarrow{R_1 – 2 R_2} \left[\begin{array}{rr|r} 1 & 0 & -2 \\ 0 & 1 & 1 \end{array} \right].
\end{align*}
With the reduced matrix, we can read off the solution $x_1 = -2$ and $x_2 = 1$.

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