MAT 350: The Row Reduction Procedure via an Example

Warm-Up Problems

Example 1: Determine the number of solutions to the linear systems corresponding to the augmented coefficient matrices in row echelon form below.

\[a)~~\left[\begin{array}{rr|r} 3 & -2 & 5\\ 0 & -5 & 10\end{array}\right]~~~~~~b) \left[\begin{array}{rr|r} 2 & 7 & -2\\ 0 & 0 & 0\end{array}\right]~~~~~~c) \left[\begin{array}{rr|r} -1 & -6 & 11\\ 0 & 0 & 3\end{array}\right]\]

\[d)~~\left[\begin{array}{rrr|r} 1 & 0 & 8 & 4\\ 0 & -6 & 2 & 1\\ 0 & 0 & 9 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]~~~~~~e) \left[\begin{array}{rrrr|r} 1 & 8 & -2 & 3 & 0\\ 0 & 0 & 2 & 0 & 1\end{array}\right]\]

Example 2: Describe the solutions to the linear systems corresponding to the augmented coefficient matrices in reduced row echelon form below.

\[a)~~\left[\begin{array}{rr|r} 1 & 0 & -3\\ 0 & 1 & 8\end{array}\right]~~~~~~b) \left[\begin{array}{rrr|r} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0\end{array}\right]~~~~~~c) \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{array}\right]\]

\[d)~~\left[\begin{array}{rrr|r} 1 & 0 & 0 & 4\\ 0 & 1 & 0 & -6\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 0 \end{array}\right]~~~~~~e) \left[\begin{array}{rrrr|r} 1 & 8 & 0 & 3 & 1\\ 0 & 0 & 1 & 0 & 4\end{array}\right]\]

Purpose

This slide deck just contains a walkthrough of a single example of solving a system via row reduction.

As we work through the example…

focus on the strategy we are utilizing
- see how we strategically work through the matrix entries
- identify how, and why, we choose particular scalars involved in our row replacement operations (\(R_i \leftarrow R_i + cR_j\))
build comfort with that row replacement operation

Note: Recall that we stated our row replacement operation as \(R_i \leftarrow R_i + cR_j\), where we replace a row with the sum of that row and a scaled copy of another row.

At times, we use an altered version of the replacement operation, \(R_i \leftarrow c_1R_i + c_2R_j\) which does row scaling and replacement all at once to save time.

Row Reduction – An Example

Example: Find all solutions to the linear system

\[\left\{\begin{array}{rcrcrcrcrcr} 2x_1 & - & x_2 & + & 3x_3 & & & + & 4x_5 & = & 10\\ x_1 & + & 2x_2 & - & 2x_3 & + & 3x_4 & - & x_5 & = & -1\\ & & x_2 & - & 4x_3 & - & x_4 & + & 2x_5 & = & 7\\ 3x_1 & - & 2x_2 & + & x_3 & + & 2x_4 & & & = & 5\\ -x_1 & + & 3x_2 & & & + & x_4 & + & 2x_5 & = & 3\end{array}\right.\]

We start by constructing the corresponding augmented coefficient matrix…

\[\left[\begin{array}{rrrrr|r} 2 & -1 & 3 & 0 & 4 & 10\\ 1 & 2 & -2 & 3 & -1 & -1\\ 0 & 1 & -4 & -1 & 2 & 7\\ 3 & -2 & 1 & 2 & 0 & 5\\ -1 & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} 2 & -1 & 3 & 0 & 4 & 10\\ 1 & 2 & -2 & 3 & -1 & -1\\ 0 & 1 & -4 & -1 & 2 & 7\\ 3 & -2 & 1 & 2 & 0 & 5\\ -1 & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{red}{2} & -1 & 3 & 0 & 4 & 10\\ \color{red}{1} & 2 & -2 & 3 & -1 & -1\\ \color{red}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{red}{2} & -1 & 3 & 0 & 4 & 10\\ \color{red}{1} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{red}{2} & -1 & 3 & 0 & 4 & 10\\ \color{red}{1} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right] \substack{R_1 \leftrightarrow R_2\\ \longrightarrow} \left[\begin{array}{rrrrr|r} \color{red}{1} & 2 & -2 & 3 & -1 & -1\\ \color{red}{2} & -1 & 3 & 0 & 4 & 10\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{1} & 2 & -2 & 3 & -1 & -1\\ \color{red}{2} & -1 & 3 & 0 & 4 & 10\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

We’ve got our pivot in column 1

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{red}{2} & -1 & 3 & 0 & 4 & 10\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

We’ve got our pivot in column 1

We’ll use it to “zero-out” all the entries underneath it!

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{red}{2} & -1 & 3 & 0 & 4 & 10\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right] \substack{R_2 \leftarrow R_2 + \left(-2\right)R_1\\ \longrightarrow} \left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{red}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

We’ve got our pivot in column 1

We’ll use it to “zero-out” all the entries underneath it!

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

We’ve got our pivot in column 1

We’ll use it to “zero-out” all the entries underneath it!

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{3} & -2 & 1 & 2 & 0 & 5\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\substack{R_4 \leftarrow R_4 + \left(-3\right)R_1\\ \longrightarrow} \left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{red}{0} & -8 & 7 & -7 & 3 & 8\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

We’ve got our pivot in column 1

We’ll use it to “zero-out” all the entries underneath it!

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{blue}{0} & -8 & 7 & -7 & 3 & 8\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

We’ve got our pivot in column 1

We’ll use it to “zero-out” all the entries underneath it!

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{blue}{0} & -8 & 7 & -7 & 3 & 8\\ \color{red}{-1} & 3 & 0 & 1 & 2 & 3\end{array}\right] \substack{R_5 \leftarrow R_5 + R_1\\ \longrightarrow} \left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{blue}{0} & -8 & 7 & -7 & 3 & 8\\ \color{red}{0} & 5 & -2 & 4 & 1 & 2\end{array}\right]\]

From here, we’ll start the row reduction procedure.

We’ll start with the leftmost column, where we want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)

We’ve got our pivot in column 1

We’ll use it to “zero-out” all the entries underneath it!

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & 2 & -2 & 3 & -1 & -1\\ \color{blue}{0} & -5 & 7 & -6 & 6 & 12\\ \color{blue}{0} & 1 & -4 & -1 & 2 & 7\\ \color{blue}{0} & -8 & 7 & -7 & 3 & 8\\ \color{blue}{0} & 5 & -2 & 4 & 1 & 2\end{array}\right]\]

Now, we’ll move to the second column

Row Reduction – An Example

\[\left[\begin{array}{rrrrr|r} \color{blue}{\boxed{1}} & \color{red}{2} & -2 & 3 & -1 & -1\\ \color{blue}{0} & \color{red}{-5} & 7 & -6 & 6 & 12\\ \color{blue}{0} & \color{red}{1} & -4 & -1 & 2 & 7\\ \color{blue}{0} & \color{red}{-8} & 7 & -7 & 3 & 8\\ \color{blue}{0} & \color{red}{5} & -2 & 4 & 1 & 2\end{array}\right]\]

Now, we’ll move to the second column, where we’ll want

A \(1\) in the pivot position (if possible, without fractions)
Every entry below that pivot to be \(0\)