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Matrix Calculations: An introduction to parts of linear algebra

Chrissy Safranski

Section 1.3 Row Operations

We’ve now seen how to solve linear systems using substitution and elimination, and how to create an augmented matrix representing a linear system. In this section, we see how to perform the variable elimination steps directly on the matrix and interpret the results to give the solution to the corresponding linear system.

Subsection Prepare

We use the method of elimination again, but directly on the augmented matrix containing only the coefficients and constants. The steps are essentially the same, except that every time we used the word “equation” before, we substitute the word “row” instead.

Example 1.3.1. Solve a system with equations and matrices side by side.

Let’s redo the work we did last section in our marbles example, this time side by side with the augmented matrix representing the system. Our first step was to write the equations with all the variables on the left side of the equal sign and in alphabetical order, as we did in Equation (1.2.1).
The words explain how we get from the current set of equations (or matrix) to the one on the next line.
\begin{equation*} \begin{array}{ccccccc} b\amp +\amp g\amp +\amp r\amp =\amp 30\\ \amp -\amp 2g\amp +\amp r\amp =\amp 0\\ -b\amp +\amp g\amp +\amp r\amp =\amp 0 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 30\\ 0\amp -2\amp 1\amp 0\\-1\amp 1\amp 1\amp 0 \end{array} \right] \end{equation*}
Replace equation 3 with the sum of equations 1 and 3.
Replace row 3 with the sum of rows 1 and 3.
\begin{equation*} \begin{array}{ccccccc} b\amp +\amp g\amp +\amp r\amp =\amp 30\\ \amp -\amp 2g\amp +\amp r\amp =\amp 0\\ \amp \amp 2g\amp +\amp 2r\amp =\amp 30 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 30\\ 0\amp -2\amp 1\amp 0\\0\amp 2\amp 2\amp 30 \end{array} \right] \end{equation*}
Multiply equation 2 by \(-\frac12\text{.}\)
Multiply row 2 by \(-\frac{1}{2}\text{.}\)
\begin{equation*} \begin{array}{ccccccc} b\amp +\amp g\amp +\amp r\amp =\amp 30\\ \amp \amp g\amp +\amp -1/2r\amp =\amp 0\\ \amp \amp 2g\amp +\amp 2r\amp =\amp 30 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 30\\ 0\amp 1\amp -\frac12\amp 0\\0\amp 2\amp 2\amp 30 \end{array} \right] \end{equation*}
Replace equation 1 with the sum of \((-1)\) times equation 2 plus equation 1. Replace equation 3 with the sum of \((-2)\) times equation 2 plus equation 3.
Replace row 1 with the sum of \((-1)\) times row 2 plus row 1. Replace row 3 with the sum of \((-2)\) times row 2 plus row 3.
\begin{equation*} \begin{array}{ccccccc} b\amp +\amp \amp \amp 3/2r\amp =\amp 30\\ \amp \amp g\amp -\amp 1/2r\amp =\amp 0\\ \amp \amp \amp \amp 3r\amp =\amp 30 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 0\amp \frac32\amp 30\\0\amp 1\amp -\frac12\amp 0\\0\amp 0\amp 3\amp 30 \end{array} \right] \end{equation*}
Multiply equation 3 by \(\frac13\text{.}\)
Multiply row 3 by \(\frac13\text{.}\)
\begin{equation*} \begin{array}{ccccccc} b\amp +\amp \amp \amp 3/2r\amp =\amp 30\\ \amp \amp g\amp -\amp 1/2r\amp =\amp 0\\ \amp \amp \amp \amp r\amp =\amp 10 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 0\amp \frac32\amp 30\\0\amp 1\amp -\frac12\amp 0\\0\amp 0\amp 1\amp 10 \end{array} \right] \end{equation*}
Replace equation 2 with the sum of \(\frac12\) times equation 3 plus equation 2. Replace equation 1 with the sum of \(-\frac32\) times equation 3 plus equation 1.
Replace row 2 with the sum of \(\frac12\) times row 3 plus row 2. Replace row 1 with the sum of \(-\frac32\) times row 3 plus row 1.
\begin{equation*} \begin{array}{ccccccc} b\amp \amp \amp \amp \amp =\amp 15\\ \amp \amp g\amp \amp \amp =\amp 5\\ \amp \amp \amp \amp r\amp =\amp 10 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 0\amp 0\amp 15\\0\amp 1\amp 0\amp 5\\0\amp 0\amp 1\amp 10 \end{array} \right] \end{equation*}
The final matrix contains the same solution information as we have on the left. Recall that the first column of our matrices held the coefficients of the \(b\) variable; the second column held the coefficients of the \(g\) variable, and the third column held the coefficients of the \(r\) variable. Therefore, the first row of the matrix can be interpreted as \(b+0g+0r=15,\) or more concisely, \(b=15.\)
This is powerful stuff! We write an augmented matrix so that we don’t have to keep writing variables over and over again, and now we can solve a linear system by performing operations on the rows of the matrix, without thinking about the linear equations at all, until we can read off the solution to the linear system at the end of the process.

Activity 1.3.2. Row operations.

In the following tasks, perform the row operations described on the given matrix.
(a)
    Consider the matrix
    \begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\text{.} \end{equation*}
    Which matrix is the result of multiplying Row 1 by \(-2\) and adding to Row 3?
  • \(\left[\begin{array}{cccc} -2\amp -2\amp 2\amp -4\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\)
  • This multiplied Row 1 by \(-2\) and replaced Row 1. It didn’t do anything to Row 3.
  • \(\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ -2\amp -7\amp -1\amp \frac{-7}{2}\\ \end{array} \right]\)
  • Correct!
  • \(\left[\begin{array}{cccc} -2\amp -2\amp 2\amp -4\\ 2\amp 1\amp 5\amp 7\\ -2\amp -7\amp -1\amp \frac{-7}{2}\\ \end{array} \right]\)
  • In the “multiply and add” row operation, you don’t change the row you multiplied by a number; you only change the row you added to.
  • \(\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ -2\amp -5\amp -3 \amp \frac{1}{2}\\ \end{array} \right]\)
  • We have to do the operation to the entire row, not just the first entry.
(b)
    Consider the matrix
    \begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\text{.} \end{equation*}
    Which matrix is the result of multiplying Row 1 by \(-2\text{?}\)
  • \(\left[\begin{array}{cccc} -2\amp -2\amp 2\amp -4\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\)
  • Correct!
  • \(\left[\begin{array}{cccc} -2\amp -2\amp 2\amp -4\\ 2\amp 1\amp 5\amp 7\\ 0\amp 10\amp 6\amp -1\\ \end{array} \right]\)
  • This multiplied both Row 1 and Row 3 by \(-2\text{.}\)
  • \(\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ -2\amp -1\amp -5\amp -7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\)
  • This multiplied Row 2 by \(-1\text{.}\)
(c)
    Consider the matrix
    \begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\text{.} \end{equation*}
    Which matrix is the result of interchanging Row 2 and Row 3?
  • \(\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ 2\amp 1\amp 5\amp 7\\ \end{array} \right]\)
  • Correct!
  • \(\left[\begin{array}{cccc} 2\amp 1\amp 5\amp 7\\ 1\amp 1\amp -1\amp 2\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\)
  • This interchanged Row 1 and Row 2.
  • \(\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right]\)
  • This replaced Row 2 with a copy of Row 3, but didn’t replace Row 3 with a copy of Row 2.
We can save ourselves even more writing by using a shorthand to describe the operations we do to the matrix rows. We use \(R_1\text{,}\) \(R_2\) to represent Row 1 and Row 2, respectively. We can write “add Row 1 to Row 3, and replace Row 3 with that sum” as “\(R_1+R_3\rightarrow R_3\text{.}\)” We also replaced Row 2 with \(\frac{1}{2}\) of Row 2; we write this in shorthand as “\(\frac{1}{2}R_2\rightarrow R_2\)”.
There is one more operation we can do to rows of a matrix without changing the corresponding system of linear equations; the solutions don’t change if we write the equations down in a different order. Thus we can swap the positions of two rows, and the expression “\(R_1 \leftrightarrow R_2\)” means “interchange Row 1 and Row 2.”

Definition 1.3.3. Elementary Row Operations.

Let \(R_i\text{,}\) \(R_j\) be two arbitrary rows of a matrix, say Row \(i\) and Row \(j\text{,}\) and let \(c\) be a nonzero number. The three elementary row operations are:
  1. Add the result of multiplying a row by a nonzero number to another row, and replace the latter row with that sum, denoted \(cR_i+R_j\rightarrow R_j\text{.}\)
  2. Multiply one row by a nonzero number, denoted \(cR_i \rightarrow R_i\text{.}\)
  3. Swap the positions of two rows, denoted \(R_i \leftrightarrow R_j\text{.}\)

Activity 1.3.4. Row operations notation.

In the following tasks, say which elementary row operation was performed on matrix \(A\) to obtain matrix \(B\text{.}\)
(a)
    Consider the matrices
    \begin{equation*} A=\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right] \end{equation*}
    \begin{equation*} B=\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 0\amp -1\amp 7\amp 3\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right] \end{equation*}
    Which row operation was done to \(A\) to obtain \(B\text{?}\)
  • \(-2R_1+R_2 \rightarrow R_2\)
  • Multiply Row 1 by \(-2\) and add to Row 2, replacing Row 2.
  • \(-2R_1+R_2 \rightarrow R_1\)
  • This says to multiply Row 1 by \(-2\text{,}\) add to Row 2, and replace Row 1. Row 1 was not changed. This is also not one of the three elementary row operations: we replace the row we added to, not the row we multiplied by a number.
  • \(2R_1-R_2 \rightarrow R_2\)
  • This would make the second row \(\begin{bmatrix} 0 \amp 1 \amp -7 \amp -3 \end{bmatrix}\text{.}\) This is also not one of the elementary row operations - we multiply and add, never multiply and subtract.
(b)
    Consider the matrices
    \begin{equation*} A=\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right] \end{equation*}
    \begin{equation*} B=\left[\begin{array}{cccc} 0\amp -5\amp -3\amp \frac{1}{2}\\ 2\amp 1\amp 5\amp 7\\ 1\amp 1\amp -1\amp 2\\ \end{array} \right] \end{equation*}
    Which row operation was done to \(A\) to obtain \(B\text{?}\)
  • \(R_1 \leftrightarrow R_3\)
  • Correct!
  • \(R_1 \leftrightarrow R_2\)
  • This says Row 1 and Row 2 were swapped.
  • \(R_1 \rightarrow R_3\)
  • This says Row 1 replaces Row 3, but we need the arrow to go both ways to say that Row 3 also replaces Row 1. This is also not one of the three elementary row operations.
(c)
    Consider the matrices
    \begin{equation*} A=\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ 0\amp -5\amp -3\amp \frac{1}{2}\\ \end{array} \right] \end{equation*}
    \begin{equation*} B=\left[\begin{array}{cccc} 1\amp 1\amp -1\amp 2\\ 2\amp 1\amp 5\amp 7\\ 0\amp -10\amp -6\amp 1\\ \end{array} \right] \end{equation*}
    Which row operation was done to \(A\) to obtain \(B\text{?}\)
  • \(2R_3 \rightarrow R_3\)
  • Correct!
  • \(-2R_3 \rightarrow R_3\)
  • This says Row 3 was multiplied by \(-2\text{.}\) Check again.
  • \(2R_3 \rightarrow R_2\)
  • This says Row 3 was multiplied by \(2\) and the result replaced Row 2. This is not one of the three elementary row operations.
Let’s see one more example of solving a linear system with the equations and the augmented matrices side by side.

Example 1.3.5. Equations and Matrices Side by Side.

Find a solution to the following system of linear equations by simultaneously manipulating the equations and the corresponding augmented matrices.
\begin{equation*} \begin{array}{ccccccc} x_1\amp +\amp x_2\amp +\amp x_3\amp =\amp 0\\ 2x_1\amp +\amp 2x_2\amp +\amp x_3\amp =\amp 0\\ -x_1\amp +\amp x_2\amp -\amp 2x_3\amp =\amp 2 \end{array} \end{equation*}
Solution.
We turn the given system into its corresponding augmented matrix.
System of equations
Corresponding matrix
\begin{equation*} \begin{array}{ccccccc} x_1\amp +\amp x_2\amp +\amp x_3\amp =\amp 0\\ 2x_1\amp +\amp 2x_2\amp +\amp x_3\amp =\amp 0\\ -x_1\amp +\amp x_2\amp -\amp 2x_3\amp =\amp 2 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 0\\ 2\amp 2\amp 1\amp 0\\ -1\amp 1\amp -2\amp 2\\ \end{array} \right] \end{equation*}
First, we use the first equation to eliminate the \(x_1\) from the second and third equations.
Replace equation 2 with the sum of \((-2)\) times equation 1 plus equation 2. Replace equation 3 with the sum of equation 1 and equation 3.
\begin{align*} -2R_1+R_2 \amp \rightarrow R_2 \\ R_1+R_3 \amp \rightarrow R_3 \end{align*}
\begin{equation*} \begin{array}{ccccccc} x_1\amp +\amp x_2\amp +\amp x_3\amp =\amp 0\\ \amp \amp \amp \amp -x_3\amp =\amp 0\\ \amp \amp 2x_2\amp -\amp x_3\amp =\amp 2 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 0\\ 0\amp 0\amp -1\amp 0\\ 0\amp 2\amp -1\amp 2\\ \end{array} \right] \end{equation*}
Notice that the second equation no longer contains \(x_2\text{.}\) We’ll exchange the order of the equations so that we can follow the convention of solving for the second variable in the second equation.
Interchange equations 2 and 3.
\begin{equation*} R_2\leftrightarrow R_3 \end{equation*}
\begin{equation*} \begin{array}{ccccccc} x_1\amp +\amp x_2\amp +\amp x_3\amp =\amp 0\\ \amp \amp 2x_2\amp -\amp x_3\amp =\amp 2\\ \amp \amp \amp \amp -x_3\amp =\amp 0 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 0\\ 0\amp 2\amp -1\amp 2\\ 0\amp 0\amp -1\amp 0\\ \end{array} \right] \end{equation*}
Multiply equation 2 by \(\frac12\text{.}\)
\begin{equation*} \frac12 R_2\rightarrow R_2 \end{equation*}
\begin{equation*} \begin{array}{ccccccc} x_1\amp +\amp x_2\amp +\amp x_3\amp =\amp 0\\ \amp \amp x_2\amp -\amp \frac12x_3\amp =\amp 1\\ \amp \amp \amp \amp -x_3\amp =\amp 0 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 0\\ 0\amp 1\amp -\frac12\amp 1\\ 0\amp 0\amp -1\amp 0\\ \end{array} \right] \end{equation*}
Multiply equation 3 by \(-1\text{.}\)
\begin{equation*} -1 R_3\rightarrow R_3 \end{equation*}
\begin{equation*} \begin{array}{ccccccc} x_1\amp +\amp x_2\amp +\amp x_3\amp =\amp 0\\ \amp \amp x_2\amp -\amp \frac12x_3\amp =\amp 1\\ \amp \amp \amp \amp x_3\amp =\amp 0 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 0\\ 0\amp 1\amp -\frac12\amp 1\\ 0\amp 0\amp 1\amp 0\\ \end{array} \right] \end{equation*}
Notice that the last row tells us that \(x_3=0\text{.}\) If we wanted to combine the methods of substitution and elimination, we could substitute \(0\) in for the \(x_3\) in the first two equations, and that would make a lot of sense! However, we choose to demonstrate how continuing with the method of elimination arrives at the same result.
Replace equation 2 with the sum of \(\frac12\) times equation 3 plus equation 2. Replace equation 1 with the sum of \((-1)\) times equation 3 plus equation 1.
\begin{align*} \frac{1}{2}R_3+R_2 \amp \rightarrow R_2 \\ -R_3+R_1 \amp \rightarrow R_1 \end{align*}
\begin{equation*} \begin{array}{ccccccc} x_1\amp +\amp x_2\amp \amp \amp =\amp 0\\ \amp \amp x_2\amp \amp \amp =\amp 1\\ \amp \amp \amp \amp x_3\amp =\amp 0 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 0\amp 0\\ 0\amp 1\amp 0\amp 1\\ 0\amp 0\amp 1\amp 0\\ \end{array} \right] \end{equation*}
Notice how the second equation shows that \(x_2 = 1\text{.}\) All that remains to do is to solve for \(x_1\text{.}\)
Replace equation 1 with the sum of \((-1)\) times equation 2 plus equation 1.
\begin{equation*} -R_2+R_1\rightarrow R_1 \end{equation*}
\begin{equation*} \begin{array}{ccccccc} x_1\amp \amp \amp \amp \amp =\amp -1\\ \amp \amp x_2\amp \amp \amp =\amp 1\\ \amp \amp \amp \amp x_3\amp =\amp 0 \end{array} \end{equation*}
\begin{equation*} \left[\begin{array}{cccc} 1\amp 0\amp 0\amp -1\\ 0\amp 1\amp 0\amp 1\\ 0\amp 0\amp 1\amp 0\\ \end{array} \right] \end{equation*}
The equations on the left tell us that \(x_1 = -1\text{,}\) \(x_2 = 1\) and \(x_3=0\text{,}\) and the matrix on the right tells us the same information.

Reading Questions Reading Questions

1. Row operation, yes or no?
    Let \(A\) be a \(4\times 5\) matrix. Which of the following notations represent elementary operations we can perform on \(A\text{?}\)
  • \(0R_1 \rightarrow R_1\)
  • We can’t multiply a row by \(0\text{,}\) only by a nonzero number. If we multiply both sides of an equation by 0, we get 0=0, which is true but rarely helpful.
  • \(5R_2+R_4 \rightarrow R_4\)
  • Replace Row 4 with the result of multiplying Row 2 by the number 5 and adding that to Row 4.
  • \(2R_1 \rightarrow R_1\)
  • Two times Row 1 replaces Row 1.
  • \(2R_1 + R_2 \rightarrow R_1\)
  • In the row operation involving multiplying and adding, we don’t replace the same row that we multiplied by a number.
  • \(R_3 \leftrightarrow R_4\)
  • Swap Row 3 and Row 4.
  • \(R_3 \rightarrow R_4 \)
  • We can’t just replace Row 4 with Row 3 without losing information about the linear system.
2. Parsons Problem, Sequence of Row Operations.
Start with matrix \(A\) and perform these row operations one after another. Give the matrix that results from doing each operation in order.
\begin{equation*} A=\begin{bmatrix} 0 \amp 1 \amp 2\amp 3 \\ 1 \amp 2\amp 3\amp 4 \\ -1 \amp -1 \amp -1 \amp -1 \end{bmatrix} \end{equation*}
\begin{align*} R_1 \amp \leftrightarrow R_2 \\ R_1+R_3 \amp \rightarrow R_3 \\ -R_2+R_3 \amp \rightarrow R_3 \end{align*}
3. Reflection.
Enter a response to both of the following tasks:
  1. Ask a question about the material, either about something you’re not sure you fully understand, or a “what if” question.
  2. Give a percentage from 0 to 100 that reflects how confident you are with the material you just read about, and give one sentence as to why you feel that way. 0 means you didn’t actually do the reading and 100 means that everything makes sense so far and you think you are completely ready to engage with the material more deeply.

Worksheet Participate
A4US

1. Row Operations Notation.

For the matrix
\begin{equation*} A=\left[\begin{array}{ccc} 2\amp -1\amp 1\\ 4\amp 0\amp 5\\ -1\amp 1\amp 1\\ \end{array} \right]\text{,} \end{equation*}
give the matrix that results from performing each of the row operations below.
(a)
\(-2R_1+R_2 \rightarrow R_2\text{.}\)
(b)
\(\frac{1}{2}R_2 \rightarrow R_2\text{.}\)
(c)
\(R_1 \leftrightarrow R_3\text{.}\)

2. Side-by-side Elimination and Matrices.

Solve the linear system below using the method of elimination side-by-side with row operations.
\begin{align*} 2x_1 + 4x_2 \amp = -6\\ 5x_1 - 5x_2 \amp = 15 \end{align*}
\begin{equation*} \begin{bmatrix} 2 \amp 4 \amp -6 \\ 5 \amp -5 \amp 15 \\ \end{bmatrix} \end{equation*}

What row operation was done?

Write down the row operation that transforms \(A\) into the matrix given in each exercise.
\begin{equation*} A=\left[\begin{array}{ccc} 2\amp -1\amp 1\\ 4\amp 0\amp 5\\ -1\amp 1\amp 1\\ \end{array} \right] \end{equation*}
3.
\begin{equation*} \left[\begin{array}{ccc} 2\amp -1\amp 1\\ 4\amp 0\amp 5\\ 2\amp -2\amp -2\\ \end{array} \right] \end{equation*}
4.
\begin{equation*} \left[\begin{array}{ccc} 0\amp 1\amp 3\\ 4\amp 0\amp 5\\ -1\amp 1\amp 1\\ \end{array} \right] \end{equation*}
5.
\begin{equation*} \left[\begin{array}{ccc} 2\amp -1\amp 1\\ -1\amp 1\amp 1\\ 4\amp 0\amp 5\\ \end{array} \right] \end{equation*}

6. Two truths and a lie?

Using shorthand row operation notation, write down two elementary row operations and one thing that looks like but isn’t an elementary row operation.

7. Exploring non-elementary row operations.

(a)
Do something to the matrix in Problem 2 that is not an elementary row operation, and write down the matrix that you get.
(b)
Solve the linear system corresponding to your new matrix. Does the linear system corresponding to your new matrix have the same solution as the linear system in Problem 2?
(c)
If the linear system has the same solution, can you write down a sequence of elementary row operations that results in your new matrix?

Summary.

  • Row operations are actions we do to the rows of a matrix that correspond to the elimination steps that we perform on the equations in a linear system.
  • Solving a linear system by performing row operations on an augmented matrix involves the same steps as elimination but with less writing.
  • There are three elementary row operations and shorthand notations for each of the three.

Subsection Practice

Exercise 1.3.1. Some row operations.

Given the matrix
\begin{equation*} A={\left[\begin{array}{cccc} -2 \amp 6 \amp 3 \amp 5\cr 0 \amp 3 \amp 5 \amp -1\cr -3 \amp -2 \amp 2 \amp -4 \end{array}\right]} \end{equation*}
perform the row operation \(R_2 \leftrightarrow R_3\) on \(A\) and enter the result.
Perform the row operation \(\frac{1}{-2}R_1 \rightarrow R_1\) on \(A\) and enter the result.

Exercise 1.3.2. A sequence of row operations.

Given the matrix
\begin{equation*} A={\left[\begin{array}{cccc} 1 \amp 1 \amp 3 \amp -2\cr 3 \amp 4 \amp 1 \amp 3\cr -3 \amp -5 \amp 4 \amp -6 \end{array}\right]}\text{,} \end{equation*}
perform the row operations below in the order given and enter the final result.
\begin{equation*} \begin{aligned} {-3} R_1 +R_2 \amp \rightarrow R_2\\ 3 R_1 +R_3 \amp \rightarrow R_3\\ 2 R_2 +R_3 \amp \rightarrow R_3 \end{aligned} \end{equation*}

Exercise 1.3.3. Identify the row operation performed.

The matrix on the right results after performing a single row operation on the matrix on the left.
\begin{equation*} {\left[\begin{array}{ccc} 1 \amp -3 \amp 1\cr -3 \amp 3 \amp 5\cr 2 \amp -4 \amp -5 \end{array}\right]} \rightarrow {\left[\begin{array}{ccc} 1 \amp -3 \amp 1\cr -9 \amp 9 \amp 15\cr 2 \amp -4 \amp -5 \end{array}\right]} \end{equation*}
Identify the row operation. Use R1, R2, and R3 to represent the rows \(R_1\text{,}\) \(R_2\text{,}\) and \(R_3\text{.}\)
Answer:
  • select
  • ->
  • <-->

Exercise 1.3.4. Identify the row operation performed.

The matrix on the right results after performing a single row operation on the matrix on the left.
\begin{equation*} {\left[\begin{array}{ccc} 2 \amp 5 \amp -6\cr -6 \amp 3 \amp -2\cr 3 \amp 1 \amp -1 \end{array}\right]} \rightarrow {\left[\begin{array}{ccc} 14 \amp 9 \amp -10\cr -6 \amp 3 \amp -2\cr 3 \amp 1 \amp -1 \end{array}\right]} \end{equation*}
Identify the row operation. Use R1, R2, and R3 to represent the rows \(R_1\text{,}\) \(R_2\text{,}\) and \(R_3\text{.}\)
Answer:
  • select
  • ->
  • <-->

Exercise 1.3.5. Identify the row operation performed.

The matrix on the right results after performing a single row operation on the matrix on the left.
\begin{equation*} {\left[\begin{array}{ccc} 4 \amp 1 \amp -1\cr 3 \amp 3 \amp 4\cr 4 \amp 6 \amp 3 \end{array}\right]} \rightarrow {\left[\begin{array}{ccc} 4 \amp 6 \amp 3\cr 3 \amp 3 \amp 4\cr 4 \amp 1 \amp -1 \end{array}\right]} \end{equation*}
Identify the row operation. Use R1, R2, and R3 to represent the rows \(R_1\text{,}\) \(R_2\text{,}\) and \(R_3\text{.}\)
Answer:
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Exercises Additional Practice

Perform row operations.

In the following exercises, perform the given row operations on \(A\text{,}\) where
\begin{equation*} A = \left[\begin{array}{ccc} 2\amp -1\amp 7\\0\amp 4\amp -2\\5\amp 0\amp 3 \end{array} \right] \end{equation*}
1.
\(-1R_1\rightarrow R_1\)
Answer.
\(\left[\begin{array}{ccc} -2\amp 1\amp -7\\0\amp 4\amp -2\\5\amp 0\amp 3\\ \end{array} \right]\)
2.
\(R_2\leftrightarrow R_3\)
Answer.
\(\left[\begin{array}{ccc} 2\amp -1\amp 7\\5\amp 0\amp 3\\0\amp 4\amp -2\\ \end{array} \right]\)
3.
\(R_1+R_2\rightarrow R_2\)
Answer.
\(\left[\begin{array}{ccc} 2\amp -1\amp 7\\2\amp 3\amp 5\\5\amp 0\amp 3\\ \end{array} \right]\)
4.
\(2R_2+R_3\rightarrow R_3\)
Answer.
\(\left[\begin{array}{ccc} 2\amp -1\amp 7\\0\amp 4\amp -2\\5\amp 8\amp -1\\ \end{array} \right]\)
5.
\(\frac12R_2\rightarrow R_2\)
Answer.
\(\left[\begin{array}{ccc} 2\amp -1\amp 7\\0\amp 2\amp -1\\5\amp 0\amp 3\\ \end{array} \right]\)
6.
\(-\frac52R_1+R_3\rightarrow R_3\)
Answer.
\(\left[\begin{array}{ccc} 2\amp -1\amp 7\\0\amp 4\amp -2\\0\amp 5/2\amp -29/2\\ \end{array} \right]\)

Which row operation was done?

A matrix \(A\) is given below. In the following exercises, a matrix \(B\) is given. Give the row operation that transforms \(A\) into \(B\text{.}\)
\begin{equation*} A = \left[\begin{array}{ccc} 1\amp 1\amp 1\\1\amp 0\amp 1\\1\amp 2\amp 3 \end{array} \right] \end{equation*}
7.
\(B = \left[\begin{array}{ccc} 1\amp 1\amp 1\\2\amp 0\amp 2\\1\amp 2\amp 3\\ \end{array} \right]\)
Answer.
\(2R_2\rightarrow R_2\)
8.
\(B = \left[\begin{array}{ccc} 1\amp 1\amp 1\\2\amp 1\amp 2\\1\amp 2\amp 3\\ \end{array} \right]\)
Answer.
\(R_1+R_2\rightarrow R_2\)
9.
\(B = \left[\begin{array}{ccc} 3\amp 5\amp 7\\1\amp 0\amp 1\\1\amp 2\amp 3\\ \end{array} \right]\)
Answer.
\(2R_3+R_1\rightarrow R_1\)
10.
\(B = \left[\begin{array}{ccc} 1\amp 0\amp 1\\1\amp 1\amp 1\\1\amp 2\amp 3\\ \end{array} \right]\)
Answer.
\(R_1\leftrightarrow R_2\)
11.
\(B = \left[\begin{array}{ccc} 1\amp 1\amp 1\\1\amp 0\amp 1\\0\amp 2\amp 2\\ \end{array} \right]\)
Answer.
\(-R_2+R_3\leftrightarrow R_3\)

Equations and matrices together.

In the following exercises, rewrite the system of equations in matrix form. Find the solution to the linear system by simultaneously manipulating the equations and the matrix.
12.
\begin{equation*} \begin{array}{ccccc} x\amp +\amp y\amp =\amp 3\\ 2x\amp -\amp 3y\amp =\amp 1 \end{array} \end{equation*}
Answer.
\(x=2,y=1\)
13.
\begin{equation*} \begin{array}{ccccc} 2x\amp +\amp 4y\amp =\amp 10\\ -x\amp +\amp y\amp =\amp 4 \end{array} \end{equation*}
Answer.
\(x=-1,y=3\)
14.
\begin{equation*} \begin{array}{ccccc} -2x\amp +\amp 3y\amp =\amp 2\\ -x\amp +\amp y\amp =\amp 1 \end{array} \end{equation*}
Answer.
\(x=-1,y=0\)
15.
\begin{equation*} \begin{array}{ccccccc} 2x\amp +\amp 3y\amp =\amp 2\\ -2x\amp +\amp 6y\amp =\amp 1 \end{array} \end{equation*}
Answer.
\(x=\frac12,y=\frac13\)
16.
\begin{equation*} \begin{array}{ccccccc} -5x_1\amp \amp \amp +\amp 2x_3\amp =\amp 14\\ \amp \amp x_2\amp \amp \amp =\amp 1\\ -3x_1\amp \amp \amp +\amp x_3\amp =\amp 8 \end{array} \end{equation*}
Answer.
\(x_1=-2,x_2=1,x_3=2\)
17.
\begin{equation*} \begin{array}{ccccccc} \amp -\amp 5x_2\amp +\amp 2x_3\amp =\amp -11\\ x_1\amp \amp \amp +\amp 2x_3\amp =\amp 15\\ \amp -\amp 3x_2\amp +\amp x_3\amp =\amp -8 \end{array} \end{equation*}
Answer.
\(x_1=1,x_2=5,x_3=7\)
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