Precalculus by Richard Wright

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# 9-02 Gaussian Elimination

Summary: In this section, you will:

• Write a matrix in reduced-row echelon form.
• Solve a system of linear equations using Gauss-Jordan Elimination.

SDA NAD Content Standards (2018): PC.6.4, PC.6.6

Even though Gaussian Elimination was not discovered by Gauss, his notation was used in Europe. In 1888, Wilhelm Jordan discovered a way to extend Gaussian Elimination, so mathematicians have named the process Gauss-Jordan Elimination. Gauss-Jordan Elimination involves using elementary row operations to write a system or equations, or matrix, in reduced-row echelon form. Reduced-row echelon form is like row echelon form, except that every element above and below and leading 1 is a 0.

###### Reduced-Row Echelon Form

A matrix is in reduced row-echelon form when

• All rows consisting entirely of zeros is at the bottom.
• For other rows the first nonzero entry is 1.
• For successive rows, the leading 1 in the higher row is further to the left.
• All entries above and below a leading 1 is a 0.

The following matrices are in reduced row-echelon form.

$$\left[\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}\right] \qquad \left[\begin{matrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]$$

#### Example 1: Identify Reduced-Row Echelon Form

Are the following matrices in reduced-row echelon form?

1. $$\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}\right]$$
2. $$\left[\begin{matrix} 1 & 2 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$$
###### Solution
1. Yes
2. No, the second row has a 1 instead of a 2. And, the first row should have a 0 in the second column. It should look something like $$\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$$
##### Try It 1

Are the following matrices in reduced-row echelon form?

1. $$\left[\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 5 \\ 0 & 0 & 0 \end{matrix}\right]$$
2. $$\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$$
1. Yes
2. Yes
###### Gauss-Jordan Elimination
1. Perform Gaussian Elimination to put the matrix in row echelon form.
2. Use elementary row operations to get zeros above each of the leading ones starting with the bottom right.
3. Continue working from the bottom up and from right to left to get zeros above each of the leading ones in each row.

#### Example 2: Put a Matrix in Reduced-Row Echelon Form

Use Gauss-Jordan Elimination to put the matrix in reduced-row echelon form.

$$\left[\begin{matrix} 1 & 2 & -2 & -6 \\ 2 & 1 & 0 & 3 \\ -1 & 3 & 3 & 4 \end{matrix}\right]$$

###### Solution

There are no zeros in the first column, so there is no need to switch any of the rows around. Start by working down the 1st column by getting rid of the 2 in the first column. Multiply the 1st row by −2 and add to 2nd row.

When showing your work on your assignment, you typically only show the steps with the red numbers in them.

$$\begin{matrix} \swarrow ×\left(-2\right) \\ +\rightarrow \qquad \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{blue}{2} & 1 & 0 & 3 \\ -1 & 3 & 3 & 4 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{red}{0} & \color{red}{-3} & \color{red}{4} & \color{red}{15} \\ -1 & 3 & 3 & 4 \end{matrix}\right]$$

Get rid of the −1 in the 1st column by adding row 1 to row 3.

$$\begin{matrix} \swarrow \\ \downarrow \quad \\ +\rightarrow \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{blue}{-1} & 3 & 3 & 4 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{red}{0} & \color{red}{5} & \color{red}{1} & \color{red}{-2} \end{matrix}\right]$$

Now work down the 2nd column. Multiply the 2nd row by 5 and add to 3 times the 3rd row.

$$\begin{matrix} \quad \\ \swarrow ×5 \\ +\rightarrow ×3 \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ 0 & \color{blue}{5} & 1 & -2 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{red}{0} & \color{red}{0} & \color{red}{23} & \color{red}{69} \end{matrix}\right]$$

The 3rd row can now be simplified, so multiply by $$\frac{1}{23}$$.

$$\begin{matrix} \quad \\ \quad \\ ×\tfrac{1}{23} \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ 0 & 0 & 23 & 69 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{red}{0} & \color{red}{0} & \color{red}{1} & \color{red}{3} \end{matrix}\right]$$

The matrix is almost in row echelon form except for the leading nonzero entry in the 2nd row is not 1. If it was turned into a 1, then there would be fractions. However, fractions would make the remaining process a bit more of a nuisance, so it will be left as a −3 for now. It is time to begin the Jordan part of the Gauss-Jordan Elimination. Because there are three rows, start getting zeros in the 3rd column and work from bottom up. Get rid of the 4 in the 2nd row by multiplying the 3rd row by −4 and adding to the 2nd row.

$$\begin{matrix} \quad \\ +\rightarrow \qquad \\ \nwarrow ×\left(-4\right) \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & \color{blue}{4} & 15 \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{red}{0} & \color{red}{-3} & \color{red}{0} & \color{red}{3} \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

The second row can be simplified, multiply it by $$-\frac{1}{3}$$.

$$\begin{matrix} \quad \\ ×\left(-\tfrac{1}{3}\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 0 & 3 \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{-1} \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

Continue working up the the 3rd column. Get rid of the −2 in the 1st row by multiplying the 3rd row by 2 and adding to the 1st row.

$$\begin{matrix} +\rightarrow\quad \\ \uparrow\qquad \\ \nwarrow ×2 \end{matrix} \left[\begin{matrix} 1 & 2 & \color{blue}{-2} & -6 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

$$\left[\begin{matrix} \color{red}{1} & \color{red}{2} & \color{red}{0} & \color{red}{0} \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

Now that the 3rd column is done, move up the 2nd column. Get rid of the 2 in the 1st row by multiplying the 2nd row by −2 and adding to the 1st row.

$$\begin{matrix} +\rightarrow \qquad \\ \nwarrow ×\left(-2\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & \color{blue}{2} & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

$$\left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{0} & \color{red}{2} \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right]$$

The matrix is now in reduced-row echelon form.

##### Try It 2

Write the matrix in reduced-row echelon form.

$$\left[\begin{matrix} 1 & 3 & -1 \\ 1 & 4 & -2 \\ -1 & -2 & 2 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}\right]$$

###### Many/No Solutions

If solving a system of linear equations with Gauss-Jordan Elimination and a row becomes all zeros with

• and the final entry is NOT zero, then no solution
• and the final entry is zero, then many solutions and use the z = a process like in lesson 8.3 example 4.

#### Example 3: Solve a System of Equations with Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

\left\{\begin{align} x + y \qquad &= 0 \\ 2x - y - z &= 5 \\ -3x + 2y + z &= -9 \end{align}\right.

###### Solution

Start by writing the system as a matrix.

$$\left[\begin{matrix} 1 & 1 & 0 & 0 \\ 2 & -1 & -1 & 5 \\ -3 & 2 & 1 & -9 \end{matrix}\right]$$

Work down the 1st column. Start by getting rid of the 2 in the 2nd row. Multiply the 1st row by −2 and add to 2nd row.

$$\begin{matrix} \swarrow ×\left(-2\right) \\ +\rightarrow \qquad \\ \quad \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{blue}{2} & -1 & -1 & 5 \\ -3 & 2 & 1 & -9 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{-3} & \color{red}{-1} & \color{red}{5} \\ -3 & 2 & 1 & -9 \end{matrix}\right]$$

Get rid of the −3 in the 1st column by multiplying the 1st row by 3 and adding it to row 3.

$$\begin{matrix} \swarrow ×3 \\ \downarrow \quad \\ +\rightarrow \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{blue}{-3} & 2 & 1 & -9 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{red}{0} & \color{red}{5} & \color{red}{1} & \color{red}{-9} \end{matrix}\right]$$

Now work down the 2nd column. Multiply the 2nd row by 5 and add to 3 times the 3rd row.

$$\begin{matrix} \quad \\ \swarrow ×5 \\ +\rightarrow ×3 \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ 0 & \color{blue}{5} & 1 & -9 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{red}{0} & \color{red}{0} & \color{red}{-2} & \color{red}{-2} \end{matrix}\right]$$

The 3rd row can now be simplified, so multiply by $$-\frac{1}{2}$$.

$$\begin{matrix} \quad \\ \quad \\ ×\left(-\tfrac{1}{2}\right) \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ 0 & 0 & -2 & -2 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{red}{0} & \color{red}{0} & \color{red}{1} & \color{red}{1} \end{matrix}\right]$$

The matrix is almost in row echelon form except for the leading nonzero entry in the 2nd row is not 1. If it was turned into a 1, then there would be fractions. However, fractions would make the remaining process a bit more of a nuisance, so it will be left as a -3 for now. It is time to begin the Jordan part of the Gauss-Jordan Elimination. Because there are three rows, start getting zeros in the 3rd column and work from bottom up. Get rid of the -1 in the 2nd row by adding the 3rd row to the 2nd row.

$$\begin{matrix} \quad \\ +\rightarrow \quad \\ \nwarrow \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & \color{blue}{-1} & 5 \\ 0 & 0 & 1 & 1 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{-3} & \color{red}{0} & \color{red}{6} \\ 0 & 0 & 1 & 1 \end{matrix}\right]$$

The second row can be simplified, multiply it by $$-\frac{1}{3}$$.

$$\begin{matrix} \quad \\ ×\left(-\tfrac{1}{3}\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & 0 & 6 \\ 0 & 0 & 1 & 1 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{-2} \\ 0 & 0 & 1 & 1 \end{matrix}\right]$$

Continue working up the the 3rd column. But there is already a zero in the 3rd column of row one. Now that the 3rd column is done, move up the 2nd column. Get rid of the 1 in the 1st row by multiplying the 2nd row by −1 and adding to the 1st row.

$$\begin{matrix} +\rightarrow \qquad \\ \nwarrow ×\left(-1\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & \color{blue}{1} & 0 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{matrix}\right]$$

$$\left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{0} & \color{red}{2} \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{matrix}\right]$$

The matrix is now in reduced-row echelon form. Remember that each row is an equation. The first row says x = 2. The second row says y = −2. And the third row says z = 1. Thus, the solution is (2, −2, 1) which also happens to be the 4th column.

##### Try It 3

Solve using Gauss-Jordan Elimination

\left\{\begin{align} 2x + y -5z &= 5 \\ y + 2z &= -1 \\ x + 3y - z &= 0 \end{align}\right.

(3, −1, 0)

#### Example 4: Solve a System of Equations with Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

\left\{\begin{align} 3x + y + z &= 10 \\ x + 2y - 3z &= 10 \\ x + y - z &= 6 \end{align}\right.

###### Solution

Start by writing the system as a matrix.

$$\left[\begin{matrix} 3 & 1 & 1 & 10 \\ 1 & 2 & -3 & 10 \\ 1 & 1 & -1 & 6 \end{matrix}\right]$$

It would be nice to have a 1 in the top left entry, so switch rows 1 and 2.

$$\left[\begin{matrix} 1 & 2 & -3 & 10 \\ 3 & 1 & 1 & 10 \\ 1 & 1 & -1 & 6 \end{matrix}\right]$$

Work down the 1st column. Start by getting rid of the 3 in the 2nd row. Multiply the 1st row by −3 and add to 2nd row.

$$\begin{matrix} \swarrow ×\left(-3\right) \\ +\rightarrow \qquad \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ \color{blue}{3} & 1 & 1 & 10 \\ 1 & 1 & -1 & 6 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -3 & 10 \\ \color{red}{0} & \color{red}{-5} & \color{red}{10} & \color{red}{-20} \\ 1 & 1 & -1 & 6 \end{matrix}\right]$$

Notice the 2nd row can be simplified by multiplying by $$-\frac{1}{5}$$.

$$\begin{matrix} \quad \\ ×\left(-\tfrac{1}{5}\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & -5 & 10 & -20 \\ 1 & 1 & -1 & 6 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -3 & 10 \\ \color{red}{0} & \color{red}{1} & \color{red}{-2} & \color{red}{4} \\ 1 & 1 & -1 & 6 \end{matrix}\right]$$

Get rid of the 1 in the 1st column by multiplying the 1st row by −1 and adding it to row 3.

$$\begin{matrix} \swarrow ×\left(-1\right) \\ \downarrow \qquad \\ +\rightarrow \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ \color{blue}{1} & 1 & -1 & 6 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ \color{red}{0} & \color{red}{-1} & \color{red}{2} & \color{red}{-4} \end{matrix}\right]$$

Now work down the 2nd column. Get rid of the −1 in the 3rd row by adding the 2nd row to the 3rd row.

$$\begin{matrix} \quad \\ \swarrow \\ +\rightarrow \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ 0 & \color{blue}{-1} & 2 & -4 \end{matrix}\right]$$

$$\left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ \color{red}{0} & \color{red}{0} & \color{red}{0} & \color{red}{0} \end{matrix}\right]$$

The 3rd row is now all zeros which means many solutions. Still finish putting the matrix in reduced-row echelon form.

It is time to begin the Jordan part of the Gauss-Jordan Elimination. Because the 2nd column has the only leading zero with a nonzero entry above it, get rid of that entry. Get rid of the 2 in the 1st row by adding the −2 times the 2nd row to the 1st row.

$$\begin{matrix} +\rightarrow \qquad \\ \nwarrow ×\left(-2\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & \color{blue}{2} & -3 & 10 \\ 0 & 1 & -2 & 4 \\ 0 & 0 & 0 & 0 \end{matrix}\right]$$

$$\left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{1} & \color{red}{2} \\ 0 & 1 & -2 & 4 \\ 0 & 0 & 0 & 0 \end{matrix}\right]$$

The matrix is now in reduced-row echelon form.

Remember that each row is an equation.

Because there are no fractions in the coefficients of z, it would be convenient to say

$$z = a$$

The second row says $$y - 2z = 4$$, so

$$y = 2a + 4$$

The first row says $$x + z = 2$$. So it becomes

$$x = -a + 2$$

Thus, the solution is $$(-a + 2, 2a + 4, a)$$.

##### Try It 4

Solve using Gauss-Jordan Elimination

\left\{\begin{align} x + 2y + 5z &= 1 \\ 3x - y - 2z &= 7 \\ 2x - 3y - 7z &= 8 \end{align}\right.

No solution

## Using a Graphing Calculator

Most graphing calculators are able to work with matrices. They can add, subtract, and multiply matrices. The graphing calculators can even put matrices into reduced-row echelon form.

###### Use a Graphing Calculator to Put a Matrix in Reduced-Row Echelon Form

TI-84

1. Press MATRIX
2. Move right to the EDIT menu and choose a matrix name
3. Set the order of the matrix and enter the elements
4. Press QUIT
5. Press MATRIX
6. Go right to the MATH menu and choose rref(
7. Press MATRIX and choose the matrix you edited
8. Close the parenthesis and press ENTER
9. NumWorks

It is faster on the NumWorks to just type in the commands instead of going through menus.

1. Press alpha twice quickly to turn on alpha lock.
2. Type rref
3. Press alpha to turn off alpha lock.
4. Press (
5. Make the matrix by pressing shift [
6. Enter all the elements. The matrix will resize as needed.
7. Move the cursor to the right outside of the matrix and press )
8. Press EXE

Or this can be done by using the menus, but it takes longer.

1. Select Calculation from the home screen.
2. Press shift [ to start a new matrix.
3. Enter the elements in the matrix. It will resize as needed.
4. Store the matrix by using the right arrow to move the cursor on the outside of the matrix, press shift alpha A EXE.
5. Press the Toolbox button and scroll down to Matrices and vectors.
6. In the new menu select Matrix.
7. Scroll down to select rref(M).
8. Press alpha A EXE
##### Lesson Summary

###### Reduced-Row Echelon Form

A matrix is in reduced row-echelon form when

• All rows consisting entirely of zeros is at the bottom.
• For other rows the first nonzero entry is 1.
• For successive rows, the leading 1 in the higher row is further to the left.
• All entries above and below a leading 1 is a 0.

The following matrices are in reduced row-echelon form.

$$\left[\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}\right] \qquad \left[\begin{matrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]$$

###### Gauss-Jordan Elimination
1. Perform Gaussian Elimination to put the matrix in row echelon form.
2. Use elementary row operations to get zeros above each of the leading ones starting with the bottom right.
3. Continue working from the bottom up and from right to left to get zeros above each of the leading ones in each row.

###### Many/No Solutions

If solving a system of linear equations with Gauss-Jordan Elimination and a row becomes all zeros with

• and the final entry is NOT zero, then no solution
• and the final entry is zero, then many solutions and use the z = a process like in lesson 8.3 example 4.

## Practice Exercises

1. What is the difference between row echelon form and reduced-row echelon form?
2. Are the following matrices in row echelon form, reduced-row echelon form, or neither?
1. $$\left[\begin{matrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}\right]$$
2. $$\left[\begin{matrix} 1 & 2 & -3 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]$$
3. $$\left[\begin{matrix} 1 & 0 & 2 & 3 \\ 0 & 1 & 1 & 0 \end{matrix}\right]$$
3. Use Gauss-Jordan Elimination to put the matrix in reduced-row echelon form.

4. $$\left[\begin{matrix} 1 & 2 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{matrix}\right]$$
5. $$\left[\begin{matrix} 1 & 2 & -3 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 2 & 4 & 1 \end{matrix}\right]$$
6. $$\left[\begin{matrix} 1 & 2 & -3 & -3 \\ -2 & 3 & 2 & -25 \\ -1 & 2 & -3 & -11 \end{matrix}\right]$$
7. Solve using Gauss-Jordan Elimination.

8. \left\{\begin{align} x + 2y - z &= -9 \\ x + y + 3z &= 10 \\ x - 2y - z &= 3 \end{align}\right.
9. \left\{\begin{align} y - 2z &= -4 \\ x + 4y - 3z &= 21 \\ -2x + y + z &= 13 \end{align}\right.
10. \left\{\begin{align} 3x + y - 5z &= 27 \\ -x + 4y + z &= -15 \\ x + 2z &= -5 \end{align}\right.
11. Use a graphing calculator to put the matrix in reduced-row echelon form.

12. $$\left[\begin{matrix} 2 & 3 & 1 \\ 1 & -4 & 2 \\ -2 & 0 & 5 \end{matrix}\right]$$
13. $$\left[\begin{matrix} 2 & 5 & -3 & -13 \\ 1 & -2 & 4 & 20 \\ -1 & 10 & 9 & 24 \end{matrix}\right]$$
14. Mixed Review

15. (9-01) What is the order of $$\left[\begin{matrix} 4 & 9 & 0 & 3 \\ 2 & 4 & 1 & 2 \end{matrix}\right]$$?
16. (8-06) Use linear programming to find the maximum of the objective function given the constraints.
Objective function: $$z = x - y$$
Constraints: \left\{\begin{align} 0 &≤ x ≤ 5 \\ y &≤ x \\ y &≥ 1 \end{align}\right.
17. (8-04) Find the partial fractions of $$\frac{4x + 14}{x^2 + 6x + 8}$$.
18. (7-09) Write the polar equation of the conic with its focus at the pole and hyperbola with eccentricity e = 2 and directrix x = −2
19. (7-08) Find the maximums of $$r = 2 \cos θ$$.

1. reduced-row echelon form is row echelon form with any entries above a leading 1 turned to zeros using elementary row operations.
2. reduced-row echelon form; row echelon form; reduced-row echelon form
3. $$\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$$
4. $$\left[\begin{matrix} 1 & 0 & -7 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]$$
5. $$\left[\begin{matrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & -1 \end{matrix}\right]$$
6. (1, −3, 4)
7. (2, 10, 7)
8. (3, −2, −4)
9. $$\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$$
10. $$\left[\begin{matrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 4 \end{matrix}\right]$$
11. 2 × 4
12. Maximum is 4 at (5, 1)
13. $$\frac{3}{x+2} + \frac{1}{x+4}$$
14. $$r = \frac{4}{1 − 2 \cos θ}$$
15. Maximums occur at $$θ = 0$$ and $$θ = π$$.