Week 4 Activity

Section 4.4 Week 4 Activity

Subsection 4.4.1 Row Reduction

Activity 4.4.1.

Row reduce this matrix.

\begin{equation*} \left( \begin{matrix} 1 \amp -3 \amp 0 \\ 1 \amp 0 \amp 7 \\ 0 \amp -4 \amp -2 \end{matrix} \right) \end{equation*}

Solution.

I proceed step-by-step through the row reduction algorithm. I’ll be very explicit about the row operation in each step. The first row already has a leading one in the first column, so I don’t need to create one. When there is a leading one, I want to clear its column. I will start by subtracting row 1 from row 2.

\begin{equation*} \left( \begin{matrix} 1 \amp -3 \amp 0 \\ 0 \amp 3 \amp 7 \\ 0 \amp -4 \amp -2 \end{matrix} \right) \end{equation*}

This has cleared the first column other than the leading 1. Now I want to create a leading one in the second column if I can. To get that, I’ll divide the second row by 3.

\begin{equation*} \left( \begin{matrix} 1 \amp -3 \amp 0 \\ 0 \amp 1 \amp \frac{7}{3} \\ 0 \amp -4 \amp -2 \end{matrix} \right) \end{equation*}

Now I have a leading 1, so I want to clear its column. I’ll add 3 times row 2 to row 1 to clear the entry above the leading 1.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 7 \\ 0 \amp 1 \amp \frac{7}{3} \\ 0 \amp -4 \amp -2 \end{matrix} \right) \end{equation*}

Then I’ll add 4 times row 2 to row 3 to clear the entry below the leading 1.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 7 \\ 0 \amp 1 \amp \frac{7}{3} \\ 0 \amp 0 \amp \frac{22}{3} \end{matrix} \right) \end{equation*}

That clears the second column. Now I want to create a leading 1 in the third column if I can. I can divide the last row by \(\frac{22}{3}\) to make a leading 1.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 7 \\ 0 \amp 1 \amp \frac{7}{3} \\ 0 \amp 0 \amp 1 \end{matrix} \right) \end{equation*}

Then I need to clear the column above this leading 1. I subtract 7 times the third row from the first row.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp \frac{7}{3} \\ 0 \amp 0 \amp 1 \end{matrix} \right) \end{equation*}

Finally, I subtract \(\frac{7}{3}\) times the third row from the second row.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{matrix} \right) \end{equation*}

This is the row reduced form.

Activity 4.4.2.

Row reduce this matrix.

\begin{equation*} \left( \begin{matrix} 4 \amp 4 \amp 0 \amp 1 \\ -1 \amp -1 \amp 2 \amp 1 \\ 6 \amp 6 \amp -4 \amp -1 \end{matrix} \right) \end{equation*}

Solution.

I proceed step-by-step through the row reduction algorithm. I’ll be very explicit about the row operation in each step.

\begin{equation*} \left( \begin{matrix} 4 \amp 4 \amp 0 \amp 1 \\ -1 \amp -1 \amp 2 \amp 1 \\ 6 \amp 6 \amp -4 \amp -1 \end{matrix} \right) \end{equation*}

First, I need a leading one in the first column. I could divide the first row by 4, but I’d like to avoid introducing fractions this early if I can. Instead, I’ll multiply the second row by \((-1)\) and then’ll interchange the first and second row.

\begin{equation*} \left( \begin{matrix} 1 \amp 1 \amp -2 \amp -1 \\ 4 \amp 4 \amp 0 \amp 1 \\ 6 \amp 6 \amp -4 \amp -1 \end{matrix} \right) \end{equation*}

Now I have a leading one in the first column. I need to clear the other entries in this column. I’ll subtract 4 times the first row from the second row.

\begin{equation*} \left( \begin{matrix} 1 \amp 1 \amp -2 \amp -1 \\ 0 \amp 0 \amp 8 \amp 5 \\ 6 \amp 6 \amp -4 \amp -1 \end{matrix} \right) \end{equation*}

Then I’ll subtract 6 times the first row from the third row.

\begin{equation*} \left( \begin{matrix} 1 \amp 1 \amp -2 \amp -1 \\ 0 \amp 0 \amp 8 \amp 5 \\ 0 \amp 0 \amp 8 \amp 5 \end{matrix} \right) \end{equation*}

Now I’ve cleared the first column. Next I’ll try to make a leading one in the second column. Looking at the matrix, though, this is impossible. The only non-zero entry in the second column is already behind a leading one -- it can never be the first non-zero entry in its row. Since it is impossible to make a leading one in the second column, I move on to the third column. The first thing I notice is that the second row and the third row are the same. So here, I’ll subtract the second row from the third row.

\begin{equation*} \left( \begin{matrix} 1 \amp 1 \amp -2 \amp -1 \\ 0 \amp 0 \amp 8 \amp 5 \\ 0 \amp 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}

Now I want to make sure the second row has a leading one in the third column and so I divide the second row by 8.

\begin{equation*} \left( \begin{matrix} 1 \amp 1 \amp -2 \amp -1 \\ 0 \amp 0 \amp 1 \amp \frac{5}{8} \\ 0 \amp 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}

Now I want to clear the third column above the leading one. To do so I will add 2 times the second row to the first row.

\begin{equation*} \left( \begin{matrix} 1 \amp 1 \amp 0 \amp \frac{1}{4} \\ 0 \amp 0 \amp 1 \amp \frac{5}{8} \\ 0 \amp 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}

This is the row reduced form.

Activity 4.4.3.

Row reduce this matrix.

\begin{equation*} \left( \begin{matrix} 2 \amp -1 \amp -3 \amp 0 \\ -6 \amp 3 \amp 5 \amp 2 \\ -4 \amp 2 \amp -2 \amp 2 \end{matrix} \right) \end{equation*}

Solution.

I want a leading one in the first row, so I’ll divide the first row by 2. (There is no avoiding fractions in this row reduction.)

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp \frac{-3}{2} \amp 0 \\ -6 \amp 3 \amp 5 \amp 2 \\ -4 \amp 2 \amp -2 \amp 2 \end{matrix} \right) \end{equation*}

Now I want to clear the first column under the leading one. I add 6 times the first row to the second.

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp \frac{-3}{2} \amp 0 \\ 0 \amp 0 \amp -4 \amp 2 \\ -4 \amp 2 \amp -2 \amp 2 \end{matrix} \right) \end{equation*}

Then I add 4 times the first row to the third.

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp \frac{-3}{2} \amp 0 \\ 0 \amp 0 \amp -4 \amp 2 \\ 0 \amp 0 \amp -8 \amp 2 \end{matrix} \right) \end{equation*}

Now the first column is finished. It is impossible to make a leading one in the second column, so I move to the third column. I’ll divide the second row by \((-4)\)

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp \frac{-3}{2} \amp 0 \\ 0 \amp 0 \amp 1 \amp \frac{-1}{2} \\ 0 \amp 0 \amp -8 \amp 2 \end{matrix} \right) \end{equation*}

Then I want to clear the third column above and below the leading one. I add \(\frac{3}{2}\) times the second row to the first row.

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp 0 \amp \frac{-3}{4} \\ 0 \amp 0 \amp 1 \amp \frac{-1}{2} \\ 0 \amp 0 \amp -8 \amp 2 \end{matrix} \right) \end{equation*}

Then I’ll add 8 times the second row to the third row.

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp 0 \amp \frac{-3}{4} \\ 0 \amp 0 \amp 1 \amp \frac{-1}{2} \\ 0 \amp 0 \amp 0 \amp -2 \end{matrix} \right) \end{equation*}

That finishes the third column, so I move to the fourth. I can make a leading one in the third row. I divide the third row by \((-2)\text{.}\)

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp 0 \amp \frac{-3}{4} \\ 0 \amp 0 \amp 1 \amp \frac{-1}{2} \\ 0 \amp 0 \amp 0 \amp 1 \end{matrix} \right) \end{equation*}

Then I want to clear the fourth column. I add \(\frac{1}{2}\) times the third row to the second row.

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp 0 \amp \frac{-3}{4} \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{matrix} \right) \end{equation*}

Finally, I add \(\frac{3}{4}\) times the third row to the first row.

\begin{equation*} \left( \begin{matrix} 1 \amp \frac{-1}{2} \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{matrix} \right) \end{equation*}

This is the reduced form.

Activity 4.4.4.

Row reduce this matrix.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp -4 \amp 2 \amp 9 \\ 0 \amp 3 \amp -2 \amp 0 \amp 7 \\ 7 \amp 0 \amp 0 \amp 2 \amp -1 \end{matrix} \right) \end{equation*}

Solution.

I proceed step-by-step through the row reduction algorithm. I’ll be very explicit about the row operation in each step.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp -4 \amp 2 \amp 9 \\ 0 \amp 3 \amp -2 \amp 0 \amp 7 \\ 7 \amp 0 \amp 0 \amp 2 \amp -1 \end{matrix} \right) \end{equation*}

There is already a leading one in the first row, and there is already a zero in the second row below it. I need to clear the first entry in the third row, so I subtract 7 times the first row from the third row.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp -4 \amp 2 \amp 9 \\ 0 \amp 3 \amp -2 \amp 0 \amp 7 \\ 0 \amp 0 \amp 28 \amp -12 \amp -64 \end{matrix} \right) \end{equation*}

There is one non-zero entry in the second column, so I’ll use that for the leading one. I divide the second row by 3.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp -4 \amp 2 \amp 9 \\ 0 \amp 1 \amp \frac{-2}{3} \amp 0 \amp \frac{7}{3} \\ 0 \amp 0 \amp 28 \amp -12 \amp -64 \end{matrix} \right) \end{equation*}

That already finishes the second column. Now I work on the third column. To make a leading one, I divide the third row by 28. (The numbers get a bit unwieldy here, but that often happens with row reduction).

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp -4 \amp 2 \amp 9 \\ 0 \amp 1 \amp \frac{-2}{3} \amp 0 \amp \frac{7}{3} \\ 0 \amp 0 \amp 1 \amp \frac{-3}{7} \amp \frac{-16}{7} \end{matrix} \right) \end{equation*}

Then I need to clear the third column above the leading one. I add \(\frac{2}{3}\) times the third row to the second row.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp -4 \amp 2 \amp 9 \\ 0 \amp 1 \amp 0 \amp \frac{-2}{7} \amp \frac{17}{21} \\ 0 \amp 0 \amp 1 \amp \frac{-3}{7} \amp \frac{-16}{7} \end{matrix} \right) \end{equation*}

Then I add 4 times the third row to the first row.

\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 0 \amp \frac{2}{7} \amp \frac{-1}{7} \\ 0 \amp 1 \amp 0 \amp \frac{-2}{7} \amp \frac{17}{21} \\ 0 \amp 0 \amp 1 \amp \frac{-3}{7} \amp \frac{-16}{7} \end{matrix} \right) \end{equation*}

This is the row reduced form.

Subsection 4.4.2 Solving Linear Systems

Activity 4.4.5.

Solve this linear system using a matrix and row reduction. Describe the solution as a locus, as a span, and in a description by free parameters.

\begin{align*} y + 3z \amp = 0 \\ -2x - 2y + z \amp = 0 \\ 4x + y - 3z \amp = 1 \end{align*}

Solution.

First, I write the matrix of the system.

\begin{equation*} \left( \begin{array}{ccc|c} 0 \amp 1 \amp 3 \amp 0 \\ -2 \amp -2 \amp 1 \amp 0 \\ 4 \amp 1 \amp -3 \amp 1 \end{array} \right) \end{equation*}

Then I row reduce this matrix. After the previous activity for row reducing matrices, I’m not going to show all the steps; I’ll just give the row reduced form calculated by a computer. You are welcome to do the same for all row reductions for the rest of the course. Here is the row reduced matrix.

\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp \frac{7}{16} \\ 0 \amp 1 \amp 0 \amp \frac{-3}{8} \\ 0 \amp 0 \amp 1 \amp \frac{1}{8} \end{array} \right) \end{equation*}

Now I need to interpret this in the three ways required by the question. The solution as a locus doesn’t need any interpretation -- the original set of three equations defined a locus since a locus is nothing but the solutions of some number of equations in the coordinate variables.

A description by parameters looks for free parameters, which are found in columns left of the dividing line without leading ones. Then I can interpret the matrix as a new system, using these free parameters. Here there are no free parameters, and there is a unique solution

\begin{align*} x \amp = \frac{7}{16}\\ y \amp = \frac{-3}{8}\\ z \amp = \frac{1}{8} \end{align*}

Finally, a description as a span is a rewriting of the description by free parameters. For every parameter, we think of the coefficients of that parameter as a vector. The vector of constants becomes the offset vector. Grouping the variables into a vector lets me write the solution as an (offset) span. Since this is a unique solution, there is only the the offset vector and no span.

\begin{equation*} \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{7}{16} \\ \frac{-3}{8} \\ \frac{1}{8} \end{matrix} \right) \end{equation*}

Activity 4.4.6.

Solve this linear system using a matrix and row reduction. Describe the solution as a locus, as a span, and in a description by free parameters.

\begin{align*} 4x - y - z \amp = 5 \\ -3x + 2z \amp = 1 \\ -2x - y + 3z \amp = 9 \end{align*}

Solution.

First I write the matrix of the system.

\begin{equation*} \left( \begin{array}{ccc|c} 4 \amp -1 \amp -1 \amp 5 \\ -3 \amp 0 \amp 2 \amp 1 \\ -2 \amp -1 \amp 3 \amp 9 \end{array} \right) \end{equation*}

Then I asked a computer for the row reduced form of the matrix.

\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp \frac{-2}{3} \amp 0 \\ 0 \amp 1 \amp \frac{-5}{3} \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{array} \right) \end{equation*}

I have a row that translates to \(0=1\text{,}\) which is a contradiction. Therefore, this system has no solutions. As there are no solutions, there is no description of those solutions, either as a locus, by parameters, or as a span.

Activity 4.4.7.

Solve this linear system using a matrix and row reduction. Describe the solution as a locus, as a span, and in a description by free parameters.

\begin{align*} -x - y + z \amp = 2 \\ 2x + 3z \amp = 1 \\ -4y + 7z \amp = 6 \end{align*}

Solution.

First, I write the matrix of the system.

\begin{equation*} \left( \begin{array}{ccc|c} -1 \amp -1 \amp 1 \amp 2 \\ 2 \amp 0 \amp 3 \amp 1 \\ 0 \amp -4 \amp 7 \amp 6 \end{array} \right) \end{equation*}

Then I asked a computer for the row reduced form of the matrix.

\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp \frac{-3}{2} \\ 0 \amp 1 \amp 0 \amp \frac{5}{6} \\ 0 \amp 0 \amp 1 \amp \frac{4}{3} \end{array} \right) \end{equation*}

A description by parameters looks for free parameters, which are found in columns left of the dividing line without leading ones. Then I can interpret the matrix as a new system, using these free parameters. Here, there are no free parameters and there is an exact solution.

\begin{align*} x \amp = \frac{-3}{2} \\ y \amp = \frac{5}{6} \\ z \amp = \frac{4}{3} \end{align*}

Finally, description as a span is a rewriting of the description by free parameters. For every parameter, I think of the coefficients of that parameter as a vector. The vector of constants becomes the offset vector. Grouping the variables into a vector lets us write the solution as an (offset) span. Since there is a unique solution, the description only includes the offset vector and no span.

\begin{equation*} \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{-3}{2} \\ \frac{5}{6} \\ \frac{4}{3} \end{matrix} \right) \end{equation*}

Activity 4.4.8.

Solve this linear system using a matrix and row reduction. Describe the solution as a locus, as a span, and in a description by free parameters.

\begin{align*} 2x - y + 6z \amp = 1 \\ 3x - 7z \amp = 1 \\ y + 4z \amp = 2 \end{align*}

Solution.

First, I write the matrix of the system.

\begin{equation*} \left( \begin{array}{ccc|c} 2 \amp -1 \amp 6 \amp 1 \\ 3 \amp 0 \amp -7 \amp 1 \\ 0 \amp 1 \amp 4 \amp 2 \end{array} \right) \end{equation*}

Then I asked a computer for the row reduced form of the matrix.

\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp \frac{31}{44} \\ 0 \amp 1 \amp 0 \amp \frac{15}{11} \\ 0 \amp 0 \amp 1 \amp \frac{7}{44} \end{array} \right) \end{equation*}

Now I need to interpret this in the three ways required by the question. The solution as a locus doesn’t need any interpretation -- the original set of three equations defined a locus, since a locus is nothing but the solutions of some number of equations in the coordinate variables.

\begin{align*} x \amp = \frac{31}{44} \\ y \amp = \frac{15}{11} \\ z \amp = \frac{7}{44} \end{align*}

Finally, description as a span is a rewriting of the description by free parameters. For every parameter, we think of the coefficients of that parameter as a vector. The vector of constants becomes the offset vector. Grouping the variables into a vector lets us write the solution as an (offset) span. Since there is a unique solution, there is only the offset vector and no span at all.

\begin{equation*} \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{31}{44} \\ \frac{15}{11} \\ \frac{7}{44} \end{matrix} \right) \end{equation*}

Activity 4.4.9.

Solve this linear system using a matrix and row reduction. Describe the solution as a locus, as a span, and in a description by free parameters.

\begin{align*} x - y + 2z \amp = 0 \\ 2x - 3z \amp = 1 \\ 4x - 2y + z \amp = 1 \\ x + 3y - 12 z \amp = 2 \end{align*}

Solution.

First, I write the matrix of the system.

\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp -1 \amp 2 \amp 0 \\ 2 \amp 0 \amp -3 \amp 1 \\ 4 \amp -2 \amp 1 \amp 1 \\ 1 \amp 3 \amp -12 \amp 2 \end{array} \right) \end{equation*}

Then I asked a computer for the row reduced form of the matrix.

\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp \frac{-3}{2} \amp \frac{1}{2} \\ 0 \amp 1 \amp \frac{-7}{2} \amp \frac{1}{2} \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}

Now I need to interpret this in the three ways required by the question. The solution as a locus doesn’t need any interpretation -- the original set of three equations defined a locus, since a locus is nothing but the solutions of some number of equations in the coordinate variables.

A description by parameters looks for free parameters, which are found in columns left of the dividing line without leading ones. Then I can interpret the matrix as a new system, using these free parameters. Here, the third column has no leading ones, so \(z\) is a free parameter.

\begin{align*} x \amp = \frac{1}{2} + \frac{3}{2} z \\ y \amp = \frac{1}{2} + \frac{7}{2} z \\ z \amp = z \end{align*}

\begin{equation*} \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{1}{2} \\ \frac{1}{2} \\ 0 \end{matrix} \right) + z \left( \begin{matrix} \frac{3}{2} \\ \frac{7}{2} \\ 1 \end{matrix} \right) = \left( \begin{matrix} \frac{1}{2} \\ \frac{1}{2} \\ 0 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} \frac{3}{2} \\ \frac{7}{2} \\ 1 \end{matrix} \right) \right\} \end{equation*}

If I want, I can scale the span to clear the fraction (since a span is all multiple, applying any multiple of the vectors in the span has no effect on the span.)

\begin{equation*} \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{1}{2} \\ \frac{1}{2} \\ 0 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} 3 \\ 7 \\ 2 \end{matrix} \right) \right\} \end{equation*}

Activity 4.4.10.

Solve this linear system using a matrix and row reduction. Describe the solution as a locus, as a span, and in a description by free parameters. In \(\RR^4\text{,}\) the order of variables that I am using in this course is \(w,x,y,z\text{.}\)

\begin{align*} w - y - z \amp = 4 \\ 5w - x + 2z \amp = 2 \\ 6x - 3y \amp = 1 \end{align*}

Solution.

First, I write the matrix of the system.

\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 0 \amp -1 \amp -1 \amp 4 \\ 5 \amp -1 \amp 0 \amp 2 \amp 2 \\ 0 \amp 6 \amp -3 \amp 0 \amp 1 \end{array} \right) \end{equation*}

Then I asked a computer for the row reduced form of the matrix.

\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 0 \amp 0 \amp \frac{5}{9} \amp \frac{1}{27} \\ 0 \amp 1 \amp 0 \amp \frac{7}{9} \amp \frac{-49}{27} \\ 0 \amp 0 \amp 1 \amp \frac{14}{9} \amp \frac{-107}{27} \end{array} \right) \end{equation*}

Now I need to interpret this in the three ways required by the question. The solution as a locus doesn’t need any interpretation -- the original set of three equations defined a locus, since a locus is nothing but the solutions of some number of equations in the coordinate variables.

A description by parameters looks for free parameters, which are found in columns left of the dividing line without leading ones. Then I can interpret the matrix as a new system, using these free parameters. Here the fourth column has no leading one, so I can use \(z\) as the free parameter.

\begin{align*} w \amp = \frac{1}{27} - \frac{5}{9} z\\ x \amp = \frac{-49}{27} - \frac{7}{9} z \\ y \amp = \frac{-107}{27} - \frac{14}{9} z \\ z \amp = z \end{align*}

\begin{equation*} \left( \begin{matrix} w \\ x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{1}{27} \\ \frac{-49}{27} \\ \frac{-107}{27} \\ 0 \end{matrix} \right) + z \left( \begin{matrix} \frac{-5}{9} \\ \frac{-7}{9} \\ \frac{-14}{9} \\ 1 \end{matrix} \right) = \left( \begin{matrix} \frac{1}{27} \\ \frac{-49}{27} \\ \frac{-107}{27} \\ 0 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} \frac{-5}{9} \\ \frac{-7}{9} \\ \frac{-14}{9} \\ 1 \end{matrix} \right) \right\} \end{equation*}

I can scale the term in the span if I want to clear the denominators.

Activity 4.4.11.

\begin{align*} -3w + 2x - z \amp = 3 \\ x + y - 4z \amp = 2 \\ 2w + x \amp = 3 \\ -w + 4x + y - 5z \amp = 8 \end{align*}

Solution.

First, I write the matrix of the system.

\begin{equation*} \left( \begin{array}{cccc|c} -3 \amp 2 \amp 0 \amp -1 \amp 3 \\ 0 \amp 1 \amp 1 \amp -4 \amp 2 \\ 2 \amp 1 \amp 0 \amp 0 \amp 3 \\ -1 \amp 4 \amp 1 \amp -5 \amp 8 \end{array} \right) \end{equation*}

Then I asked a computer for the row reduced form of the matrix.

\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 0 \amp 0 \amp \frac{1}{7} \amp \frac{3}{7} \\ 0 \amp 1 \amp 0 \amp \frac{-2}{7} \amp \frac{15}{7} \\ 0 \amp 0 \amp 1 \amp \frac{-26}{7} \amp \frac{-1}{7} \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}

Now I need to interpret this in the three ways required by the question. The solution as a locus doesn’t need any interpretation -- the original set of three equations defined a locus, since a locus is nothing but the solutions of some number of equations in the coordinate variables.

A description by parameters looks for free parameters, which are found in columns left of the dividing line without leading ones. Then I can interpret the matrix as a new system, using these free parameters. Since the fourth column has no leading one, I can use \(z\) as a free parameter.

\begin{align*} w \amp = \frac{3}{7} - \frac{1}{7} z \\ x \amp = \frac{15}{7} + \frac{2}{7} z \\ y \amp = \frac{-1}{7} - \frac{26}{7} z \\ z \amp = z \end{align*}

\begin{equation*} \left( \begin{matrix} w \\ x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{3}{7} \\ \frac{15}{7} \\ \frac{-1}{7} \\ 0 \end{matrix} \right) + z \left( \begin{matrix} \frac{-1}{7} \\ \frac{2}{7} \\ \frac{-26}{7} \\ 1 \end{matrix} \right) = \left( \begin{matrix} \frac{3}{7} \\ \frac{15}{7} \\ \frac{-1}{7} \\ 0 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} \frac{-1}{7} \\ \frac{2}{7} \\ \frac{-26}{7} \\ 1 \end{matrix} \right) \right\} \end{equation*}

I can scale the vector in the span if I want to clear denominators.

\begin{equation*} \left( \begin{matrix} w \\ x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} \frac{3}{7} \\ \frac{15}{7} \\ \frac{-1}{7} \\ 0 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} -1 \\ 2 \\ -26 \\ 7 \end{matrix} \right) \right\} \end{equation*}

Subsection 4.4.3 Conceptual Review Questions

What is a linear system?
What does a matrix encode a linear system?
What is row reduction?
What is description by parameters?

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