2.2 Systems of Linear Equations; Gauss-Jordan Method

In this section, you will learn to:
  1. Represent a system of linear equations as an augmented matrix
  2. Solve the system using elementary row operations.

In this section, we learn to solve systems of linear equations using a process called the Gauss-Jordan method. The process begins by first expressing the system as a matrix, and then reducing it to an equivalent system by simple row operations. The process is continued until the solution is obvious from the matrix. The matrix that represents the system is called the augmented matrix, and the arithmetic manipulation that is used to move from a system to a reduced equivalent system is called a row operation.

Example 2.2.1

Write the following system as an augmented matrix.

\[ \begin{array}{l} 2 x + 3 y - 4 z = 5 \\ 3 x + 4 y - 5 z = - 6 \\ 4 x + 5 y - 6 z = 7 \\ \end{array} \]
Example 2.2.1 Solution

We express the above information in matrix form. Since a system is entirely determined by its coefficient matrix and by its matrix of constant terms, the augmented matrix will include only the coefficient matrix and the constant matrix. So the augmented matrix we get is as follows:

\[ \left[ \begin{array}{c c c c c} 2 & 3 & - 4 & | & 5 \\ 3 & 4 & - 5 & | & - 6 \\ 4 & 5 & - 6 & | & 7 \end{array} \right] \]

In the last section, we expressed the system of equations as \(\mathrm{AX} = \mathrm{B}\), where A represented the coefficient matrix, and B the matrix of constant terms. As an augmented matrix, we write the matrix as [A | B]. It is clear that all of the information is maintained in this matrix form, and only the letters x, y and z are missing. A student may choose to write x, y and z on top of the first three columns to help ease the transition.

Example 2.2.2

For the following augmented matrix, write the system of equations it represents.

\[ \left[ \begin{array}{c c c c c} 1 & 3 & - 5 & | & 2 \\ 2 & 0 & - 3 & | & - 5 \\ 3 & 2 & - 3 & | & - 1 \end{array} \right] \]
Example 2.2.2 Solution

The system is readily obtained as below.

\[ \begin{array}{l} \mathrm {x} + 3 \mathrm {y} - 5 \mathrm {z} = 2 \\ 2 x \quad - 3 z = - 5 \\ 3 x + 2 y - 3 z = - 1 \\ \end{array} \]

Once a system is expressed as an augmented matrix, the Gauss-Jordan method reduces the system into a series of equivalent systems by using the row operations. This row reduction continues until the system is expressed in what is called the reduced row echelon form.

The reduced row echelon form of the coefficient matrix has 1's along the main diagonal and zeros elsewhere. The solution is readily obtained from this form.

The method is not much different from the algebraic operations we employed in the elimination method in the first chapter. The basic difference is that it is algorithmic in nature, and, therefore, can easily be programmed on a computer.

We will next solve a system of two equations with two unknowns, using the elimination method, and then show that the method is analogous to the Gauss-Jordan method.

Example 2.2.3

Solve the following system by the elimination method.

\[ \mathrm {x} + 3 \mathrm {y} = 7 \] \[ 3 x + 4 y = 1 1 \]
Example 2.2.3 Solution

We multiply the first equation by \(-3\), and add it to the second equation.

\[ \begin{array}{l} - 3 x - 9 y = - 2 1 \\ \frac {3 x + 4 y = 1 1}{- 5 y = - 1 0} \\ \end{array} \]

By doing this we transformed our original system into an equivalent system:

\[ \begin{array}{l} \mathrm {x} + 3 \mathrm {y} = 7 \\ - 5 \mathrm {y} = - 1 0 \\ \end{array} \]

We divide the second equation by \(-5\), and we get the next equivalent system.

\[ \begin{array}{l} \mathrm {x} + 3 \mathrm {y} = 7 \\ \mathrm {y} = 2 \\ \end{array} \]

Now we multiply the second equation by \(-3\) and add to the first, we get

\[ \begin{array}{l} \mathrm {x} = 1 \\ \mathrm {y} = 2 \\ \end{array} \]
Example 2.2.4

Solve the following system from Example 2.2.3 by the Gauss-Jordan method, and show the similarities in both methods by writing the equations next to the matrices.

\[ \begin{array}{l} \mathrm {x} + 3 \mathrm {y} = 7 \\ 3 x + 4 y = 1 1 \\ \end{array} \]
Example 2.2.4 Solution

The augmented matrix for the system is as follows.

\[ \begin{array}{l} \left[ \begin{array}{c c c c} 1 & 3 & | & 7 \\ 3 & 4 & | & 1 1 \end{array} \right] \\ \left[ \begin{array}{l l} \mathrm {x} + 3 \mathrm {y} & = 7 \\ 3 \mathrm {x} + 4 \mathrm {y} & = 1 1 \end{array} \right] \\ \end{array} \]

We multiply the first row by \(-3\), and add to the second row.

\[ \begin{array}{l} \left[ \begin{array}{c c c c} 1 & 3 & & 7 \\ 0 & - 5 & & - 1 0 \end{array} \right] \\ \left[ \begin{array}{r l} & \mathrm {x} + 3 \mathrm {y} = 7 \\ & - 5 \mathrm {y} = - 1 0 \end{array} \right] \\ \end{array} \]

We divide the second row by \(-5\), we get,

\[ \begin{array}{l} \left[ \begin{array}{c c c c} 1 & 3 & 7 \\ 0 & 1 & 2 \end{array} \right] \\ \left[ \begin{array}{r l} \mathrm {x} + 3 \mathrm {y} & = 7 \\ \mathrm {y} & = 2 \end{array} \right] \\ \end{array} \]

Finally, we multiply the second row by \(-3\) and add to the first row, and we get,

\[ \left[ \begin{array}{c c c c} 1 & 0 & | & 1 \\ 0 & 1 & | & 2 \end{array} \right] \] \[ \left[ \begin{array}{l l} \mathrm {x} & = 1 \\ \mathrm {y} & = 2 \end{array} \right] \]

Now we list the three row operations the Gauss-Jordan method employs.

2.2.1 Row Operations

The Three Elementary Row Operations
  1. Any two rows in the augmented matrix may be interchanged.
  2. Any row may be multiplied by a non-zero constant.
  3. A constant multiple of a row may be added to another row.

One can easily see that these three row operations may make the system look different, but they do not change the solution of the system.

Row Operation 1: Row Interchange

The first row operation states that if any two rows of a system are interchanged, the new system obtained has the same solution as the old one. Let us look at an example in two equations with two unknowns. Consider the system

\[ \mathrm {x} + 3 \mathrm {y} = 7 \] \[ 3 x + 4 y = 1 1 \]

We interchange the rows, and we get,

\[ 3 x + 4 y = 1 1 \] \[ \mathrm {x} + 3 \mathrm {y} = 7 \]

Clearly, this system has the same solution as the one above.

Try It Now — Row Interchange

Given the augmented matrix:

\[\left[\begin{array}{c c c c} 0 & 2 & | & 6 \\ 1 & -3 & | & 4 \end{array}\right]\]

Perform a row interchange so that the first row, first column entry is nonzero. Write the resulting matrix.

Solution

Interchange R1 and R2:

\[\left[\begin{array}{c c c c} 1 & -3 & | & 4 \\ 0 & 2 & | & 6 \end{array}\right]\]

Now the first entry in row 1 is 1 (nonzero), which positions us to continue with elimination.

Row Operation 2: Scalar Multiplication

The second row operation states that if a row is multiplied by any non-zero constant, the new system obtained has the same solution as the old one. Consider the above system again,

\[ \mathrm {x} + 3 \mathrm {y} = 7 \] \[ 3 x + 4 y = 1 1 \]

We multiply the first row by \(-3\), we get,

\[ - 3 x - 9 y = - 2 1 \] \[ 3 x + 4 y = 1 1 \]

Again, it is obvious that this new system has the same solution as the original.

Try It Now — Scalar Multiplication

Given the augmented matrix:

\[\left[\begin{array}{c c c c} 1 & 4 & | & 10 \\ 0 & -3 & | & 9 \end{array}\right]\]

Multiply the second row by a constant so that the leading entry becomes 1. Write the resulting matrix.

Solution

Divide R2 by \(-3\) (equivalently, multiply by \(-\tfrac{1}{3}\)):

\[\left[\begin{array}{c c c c} 1 & 4 & | & 10 \\ 0 & 1 & | & -3 \end{array}\right]\]

The leading entry in row 2 is now 1.

Row Operation 3: Row Addition

The third row operation states that any constant multiple of one row added to another preserves the solution. Consider our system,

\[ \mathrm {x} + 3 \mathrm {y} = 7 \] \[ 3 x + 4 y = 1 1 \]

If we multiply the first row by \(-3\), and add it to the second row, we get,

\[ \mathrm {x} + 3 \mathrm {y} = 7 \] \[ - 5 y = - 1 0 \]

And once again, the same solution is maintained.

Try It Now — Row Addition

Given the augmented matrix:

\[\left[\begin{array}{c c c c} 1 & 5 & | & 12 \\ 3 & 2 & | & 7 \end{array}\right]\]

Use a row operation to make the entry in row 2, column 1 a zero. Write the resulting matrix and state which operation you used.

Solution

Multiply R1 by \(-3\) and add to R2: \(\quad R_2 \leftarrow R_2 - 3R_1\)

\[\left[\begin{array}{c c c c} 1 & 5 & | & 12 \\ 0 & -13 & | & -29 \end{array}\right]\]

The entry in row 2, column 1 is now 0, while row 1 is unchanged.

Now that we understand how the three row operations work, it is time to introduce the Gauss-Jordan method to solve systems of linear equations.

As mentioned earlier, the Gauss-Jordan method starts out with an augmented matrix, and by a series of row operations ends up with a matrix that is in the reduced row echelon form.

Reduced Row Echelon Form (RREF)

A matrix is in reduced row echelon form if it satisfies ALL of the following:

  1. The first nonzero entry in each row is a 1 (called a leading 1)
  2. The columns containing leading 1's have all other entries as zeros
  3. The leading entry in each row is to the right of the leading entry in the row above it
  4. Rows containing all zeros are at the bottom of the matrix

Try It Now — Identifying Reduced Row Echelon Form

Determine whether each matrix is in reduced row echelon form. If not, state which requirement(s) it violates.

(a)

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & -2 \\ 0 & 0 & 1 & | & 5 \end{array}\right]\]

(b)

\[\left[\begin{array}{c c c c c} 1 & 2 & 0 & | & 4 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & 6 \end{array}\right]\]

(c)

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 7 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 1 & | & 2 \end{array}\right]\]
Solution

(a) YES — This matrix IS in reduced row echelon form. It satisfies all four requirements: each leading entry is 1, columns with leading 1's have zeros elsewhere, leading 1's move to the right, and there are no all-zero rows.

(b) NO — Violates requirement 2. The column containing the leading 1 in row 2 (column 2) has a nonzero entry (2) in row 1. For RREF, all other entries in that column must be zero.

(c) NO — Violates requirement 4. The all-zero row (row 2) should be at the bottom, but it appears above row 3 which contains a leading 1.

We state the Gauss-Jordan method as follows.

2.2.2 Gauss-Jordan Method

The Gauss-Jordan Method
  1. Write the augmented matrix.
  2. Interchange rows if necessary to obtain a non-zero number in the first row, first column.
  3. Use a row operation to get a 1 as the entry in the first row and first column.
  4. Use row operations to make all other entries zeros in column one.
  5. Interchange rows if necessary to obtain a nonzero number in the second row, second column. Use a row operation to make this entry 1. Use row operations to make all other entries zeros in column two.
  6. Repeat step 5 for row 3, column 3. Continue moving along the main diagonal until you reach the last row, or until the number is zero.
  7. The final matrix is called the reduced row-echelon form.

Try It Now — Gauss-Jordan Method

Use the Gauss-Jordan method to solve the following system. Show each step.

\[\begin{cases} x + 2y = 5 \\ 3x + 5y = 14 \end{cases}\]
Solution

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c} 1 & 2 & | & 5 \\ 3 & 5 & | & 14 \end{array}\right]\]

Step 2: Eliminate the entry in row 2, column 1. Multiply row 1 by \(-3\) and add to row 2.

\[\left[\begin{array}{c c c c} 1 & 2 & | & 5 \\ 0 & -1 & | & -1 \end{array}\right] \quad R_2 \leftarrow R_2 - 3R_1\]

Step 3: Make the pivot in row 2, column 2 equal to 1. Divide row 2 by \(-1\).

\[\left[\begin{array}{c c c c} 1 & 2 & | & 5 \\ 0 & 1 & | & 1 \end{array}\right] \quad R_2 \leftarrow \frac{R_2}{-1}\]

Step 4: Eliminate the entry in row 1, column 2. Multiply row 2 by \(-2\) and add to row 1.

\[\left[\begin{array}{c c c c} 1 & 0 & | & 3 \\ 0 & 1 & | & 1 \end{array}\right] \quad R_1 \leftarrow R_1 - 2R_2\]

Answer: \(x = 3\), \(y = 1\)

Verification: \(3 + 2(1) = 5\) ✓ and \(3(3) + 5(1) = 14\) ✓

Example 2.2.5

Solve the following system by the Gauss-Jordan method.

\[ 2 x + y + 2 z = 1 0 \] \[ \mathrm {x} + 2 \mathrm {y} + \mathrm {z} = 8 \] \[ 3 x + y - z = 2 \]
Example 2.2.5 Solution

We write the augmented matrix.

\[ \left[ \begin{array}{c c c c c} 2 & 1 & 2 & | & 1 0 \\ 1 & 2 & 1 & | & 8 \\ 3 & 1 & - 1 & | & 2 \end{array} \right] \]

We want a 1 in row one, column one. This can be obtained by dividing the first row by 2, or interchanging the second row with the first. Interchanging the rows is a better choice because that way we avoid fractions.

\[ \left[ \begin{array}{c c c c c} 1 & 2 & 1 & | & 8 \\ 2 & 1 & 2 & | & 1 0 \\ 3 & 1 & - 1 & | & 2 \end{array} \right] \]

we interchanged row 1(R1) and row 2(R2)

We need to make all other entries zeros in column 1. To make the entry (2) a zero in row 2, column 1, we multiply row 1 by -2 and add it to the second row. We get,

\[ \left[ \begin{array}{r r r r r} 1 & 2 & 1 & | & 8 \\ 0 & - 3 & 0 & | & - 6 \\ 3 & 1 & - 1 & | & 2 \end{array} \right] \quad - 2 \mathrm {R} 1 + \mathrm {R} 2 \]

To make the entry (3) a zero in row 3, column 1, we multiply row 1 by -3 and add it to the third row. We get,

\[ \left[ \begin{array}{r r r r r} 1 & 2 & 1 & | & 8 \\ 0 & - 3 & 0 & | & - 6 \\ 0 & - 5 & - 4 & | & - 2 2 \end{array} \right] \quad - 3 \mathrm {R} 1 + \mathrm {R} 3 \]

So far we have made a 1 in the left corner and all other entries zeros in that column. Now we move to the next diagonal entry, row 2, column 2. We need to make this entry(-3) a 1 and make all other entries in this column zeros. To make row 2, column 2 entry a 1, we divide the entire second row by -3.

\[ \left[ \begin{array}{c c c c c} 1 & 2 & 1 & | & 8 \\ 0 & 1 & 0 & | & 2 \\ 0 & - 5 & - 4 & | & - 2 2 \end{array} \right] \quad \mathrm {R} 2 \div (- 3) \]

Next, we make all other entries zeros in the second column.

\[ \left[ \begin{array}{c c c c c} 1 & 0 & 1 & | & 4 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & - 4 & | & - 1 2 \end{array} \right] \quad - 2 R 2 + R 1 \text {a n d} 5 R 2 + R 3 \]

We make the last diagonal entry a 1, by dividing row 3 by -4.

\[ \left[ \begin{array}{c c c c c} 1 & 0 & 1 & | & 4 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 3 \end{array} \right] \quad \mathrm {R} 3 \div (- 4) \]

Finally, we make all other entries zeros in column 3.

\[ \left[ \begin{array}{c c c c c} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 3 \end{array} \right] \quad - R 3 + R 1 \]

Clearly, the solution reads \(x = 1\), \(y = 2\), and \(z = 3\).

Before we leave this section, we mention some terms we may need in the fourth chapter.

Pivoting Terminology

The process of obtaining a 1 in a location, and then making all other entries zeros in that column, is called pivoting.

The number that is made a 1 is called the pivot element, and the row that contains the pivot element is called the pivot row.

We often multiply the pivot row by a number and add it to another row to obtain a zero in the latter. The row to which a multiple of pivot row is added is called the target row.

Problem Set 2.2

Solve the following by the Gauss-Jordan Method. Show all work.

1. \(\begin{cases} x + 3y = 1 \\ 2x - 5y = 13 \end{cases}\)

Problem 1 Solution

Solve the system:

\[x + 3y = 1\] \[2x - 5y = 13\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c} 1 & 3 & | & 1 \\ 2 & -5 & | & 13 \end{array}\right]\]

Step 2: Eliminate the entry in row 2, column 1. Multiply row 1 by \(-2\) and add to row 2.

\[\left[\begin{array}{c c c c} 1 & 3 & | & 1 \\ 0 & -11 & | & 11 \end{array}\right] \quad R_2 \leftarrow R_2 - 2R_1\]

Step 3: Make the pivot in row 2, column 2 equal to 1. Divide row 2 by \(-11\).

\[\left[\begin{array}{c c c c} 1 & 3 & | & 1 \\ 0 & 1 & | & -1 \end{array}\right] \quad R_2 \leftarrow \frac{R_2}{-11}\]

Step 4: Eliminate the entry in row 1, column 2. Multiply row 2 by \(-3\) and add to row 1.

\[\left[\begin{array}{c c c c} 1 & 0 & | & 4 \\ 0 & 1 & | & -1 \end{array}\right] \quad R_1 \leftarrow R_1 - 3R_2\]

Answer: \(x = 4\), \(y = -1\)

Verification: \(4 + 3(-1) = 4 - 3 = 1\) ✓ and \(2(4) - 5(-1) = 8 + 5 = 13\) ✓

2. \(\begin{cases} x - y - z = -1 \\ x - 3y + 2z = 7 \\ 2x - y + z = 3 \end{cases}\)

Problem 2 Solution

Solve the system:

\[x - y - z = -1\] \[x - 3y + 2z = 7\] \[2x - y + z = 3\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c c} 1 & -1 & -1 & | & -1 \\ 1 & -3 & 2 & | & 7 \\ 2 & -1 & 1 & | & 3 \end{array}\right]\]

Step 2: Eliminate the entry in row 2, column 1. Subtract row 1 from row 2.

\[\left[\begin{array}{c c c c c} 1 & -1 & -1 & | & -1 \\ 0 & -2 & 3 & | & 8 \\ 2 & -1 & 1 & | & 3 \end{array}\right] \quad R_2 \leftarrow R_2 - R_1\]

Step 3: Eliminate the entry in row 3, column 1. Multiply row 1 by \(-2\) and add to row 3.

\[\left[\begin{array}{c c c c c} 1 & -1 & -1 & | & -1 \\ 0 & -2 & 3 & | & 8 \\ 0 & 1 & 3 & | & 5 \end{array}\right] \quad R_3 \leftarrow R_3 - 2R_1\]

Step 4: Make the pivot in row 2, column 2 equal to 1. Divide row 2 by \(-2\).

\[\left[\begin{array}{c c c c c} 1 & -1 & -1 & | & -1 \\ 0 & 1 & -\tfrac{3}{2} & | & -4 \\ 0 & 1 & 3 & | & 5 \end{array}\right] \quad R_2 \leftarrow \frac{R_2}{-2}\]

Step 5: Eliminate the entry in row 1, column 2. Add row 2 to row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & -\tfrac{5}{2} & | & -5 \\ 0 & 1 & -\tfrac{3}{2} & | & -4 \\ 0 & 1 & 3 & | & 5 \end{array}\right] \quad R_1 \leftarrow R_1 + R_2\]

Step 6: Eliminate the entry in row 3, column 2. Subtract row 2 from row 3.

\[\left[\begin{array}{c c c c c} 1 & 0 & -\tfrac{5}{2} & | & -5 \\ 0 & 1 & -\tfrac{3}{2} & | & -4 \\ 0 & 0 & \tfrac{9}{2} & | & 9 \end{array}\right] \quad R_3 \leftarrow R_3 - R_2\]

Step 7: Make the pivot in row 3, column 3 equal to 1. Multiply row 3 by \(\tfrac{2}{9}\).

\[\left[\begin{array}{c c c c c} 1 & 0 & -\tfrac{5}{2} & | & -5 \\ 0 & 1 & -\tfrac{3}{2} & | & -4 \\ 0 & 0 & 1 & | & 2 \end{array}\right] \quad R_3 \leftarrow \frac{2}{9} R_3\]

Step 8: Eliminate the entry in row 1, column 3. Add \(\tfrac{5}{2}\) times row 3 to row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 0 \\ 0 & 1 & -\tfrac{3}{2} & | & -4 \\ 0 & 0 & 1 & | & 2 \end{array}\right] \quad R_1 \leftarrow R_1 + \tfrac{5}{2}R_3\]

Step 9: Eliminate the entry in row 2, column 3. Add \(\tfrac{3}{2}\) times row 3 to row 2.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & | & -1 \\ 0 & 0 & 1 & | & 2 \end{array}\right] \quad R_2 \leftarrow R_2 + \tfrac{3}{2}R_3\]

Answer: \(x = 0\), \(y = -1\), \(z = 2\)

Verification: \(0 - (-1) - 2 = -1\) ✓, \(0 - 3(-1) + 2(2) = 3 + 4 = 7\) ✓, and \(2(0) - (-1) + 2 = 3\) ✓

3. \(\begin{cases} x + 2y + 3z = 9 \\ 3x + 4y + z = 5 \\ 2x - y + 2z = 11 \end{cases}\)

Problem 3 Solution

Solve the system:

\[x + 2y + 3z = 9\] \[3x + 4y + z = 5\] \[2x - y + 2z = 11\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c c} 1 & 2 & 3 & | & 9 \\ 3 & 4 & 1 & | & 5 \\ 2 & -1 & 2 & | & 11 \end{array}\right]\]

Step 2: Eliminate the entry in row 2, column 1. Multiply row 1 by \(-3\) and add to row 2.

\[\left[\begin{array}{c c c c c} 1 & 2 & 3 & | & 9 \\ 0 & -2 & -8 & | & -22 \\ 2 & -1 & 2 & | & 11 \end{array}\right] \quad R_2 \leftarrow R_2 - 3R_1\]

Step 3: Eliminate the entry in row 3, column 1. Multiply row 1 by \(-2\) and add to row 3.

\[\left[\begin{array}{c c c c c} 1 & 2 & 3 & | & 9 \\ 0 & -2 & -8 & | & -22 \\ 0 & -5 & -4 & | & -7 \end{array}\right] \quad R_3 \leftarrow R_3 - 2R_1\]

Step 4: Make the pivot in row 2, column 2 equal to 1. Divide row 2 by \(-2\).

\[\left[\begin{array}{c c c c c} 1 & 2 & 3 & | & 9 \\ 0 & 1 & 4 & | & 11 \\ 0 & -5 & -4 & | & -7 \end{array}\right] \quad R_2 \leftarrow \frac{R_2}{-2}\]

Step 5: Eliminate the entry in row 1, column 2. Multiply row 2 by \(-2\) and add to row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & -5 & | & -13 \\ 0 & 1 & 4 & | & 11 \\ 0 & -5 & -4 & | & -7 \end{array}\right] \quad R_1 \leftarrow R_1 - 2R_2\]

Step 6: Eliminate the entry in row 3, column 2. Add 5 times row 2 to row 3.

\[\left[\begin{array}{c c c c c} 1 & 0 & -5 & | & -13 \\ 0 & 1 & 4 & | & 11 \\ 0 & 0 & 16 & | & 48 \end{array}\right] \quad R_3 \leftarrow R_3 + 5R_2\]

Step 7: Make the pivot in row 3, column 3 equal to 1. Divide row 3 by 16.

\[\left[\begin{array}{c c c c c} 1 & 0 & -5 & | & -13 \\ 0 & 1 & 4 & | & 11 \\ 0 & 0 & 1 & | & 3 \end{array}\right] \quad R_3 \leftarrow \frac{R_3}{16}\]

Step 8: Eliminate the entry in row 1, column 3. Add 5 times row 3 to row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 2 \\ 0 & 1 & 4 & | & 11 \\ 0 & 0 & 1 & | & 3 \end{array}\right] \quad R_1 \leftarrow R_1 + 5R_3\]

Step 9: Eliminate the entry in row 2, column 3. Multiply row 3 by \(-4\) and add to row 2.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & -1 \\ 0 & 0 & 1 & | & 3 \end{array}\right] \quad R_2 \leftarrow R_2 - 4R_3\]

Answer: \(x = 2\), \(y = -1\), \(z = 3\)

Verification: \(2 + 2(-1) + 3(3) = 2 - 2 + 9 = 9\) ✓, \(3(2) + 4(-1) + 3 = 6 - 4 + 3 = 5\) ✓, and \(2(2) - (-1) + 2(3) = 4 + 1 + 6 = 11\) ✓

4. \(\begin{cases} x + 2y = 0 \\ y + z = 3 \\ x + 3z = 14 \end{cases}\)

Problem 4 Solution

Solve the system:

\[x + 2y = 0\] \[y + z = 3\] \[x + 3z = 14\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c c} 1 & 2 & 0 & | & 0 \\ 0 & 1 & 1 & | & 3 \\ 1 & 0 & 3 & | & 14 \end{array}\right]\]

Step 2: Eliminate the entry in row 3, column 1. Subtract row 1 from row 3.

\[\left[\begin{array}{c c c c c} 1 & 2 & 0 & | & 0 \\ 0 & 1 & 1 & | & 3 \\ 0 & -2 & 3 & | & 14 \end{array}\right] \quad R_3 \leftarrow R_3 - R_1\]

Step 3: Eliminate the entry in row 1, column 2. Multiply row 2 by \(-2\) and add to row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & -2 & | & -6 \\ 0 & 1 & 1 & | & 3 \\ 0 & -2 & 3 & | & 14 \end{array}\right] \quad R_1 \leftarrow R_1 - 2R_2\]

Step 4: Eliminate the entry in row 3, column 2. Add 2 times row 2 to row 3.

\[\left[\begin{array}{c c c c c} 1 & 0 & -2 & | & -6 \\ 0 & 1 & 1 & | & 3 \\ 0 & 0 & 5 & | & 20 \end{array}\right] \quad R_3 \leftarrow R_3 + 2R_2\]

Step 5: Make the pivot in row 3, column 3 equal to 1. Divide row 3 by 5.

\[\left[\begin{array}{c c c c c} 1 & 0 & -2 & | & -6 \\ 0 & 1 & 1 & | & 3 \\ 0 & 0 & 1 & | & 4 \end{array}\right] \quad R_3 \leftarrow \frac{R_3}{5}\]

Step 6: Eliminate the entry in row 1, column 3. Add 2 times row 3 to row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 2 \\ 0 & 1 & 1 & | & 3 \\ 0 & 0 & 1 & | & 4 \end{array}\right] \quad R_1 \leftarrow R_1 + 2R_3\]

Step 7: Eliminate the entry in row 2, column 3. Subtract row 3 from row 2.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & -1 \\ 0 & 0 & 1 & | & 4 \end{array}\right] \quad R_2 \leftarrow R_2 - R_3\]

Answer: \(x = 2\), \(y = -1\), \(z = 4\)

Verification: \(2 + 2(-1) = 0\) ✓, \(-1 + 4 = 3\) ✓, and \(2 + 3(4) = 2 + 12 = 14\) ✓

5. Two apples and four bananas cost $2.00 and three apples and five bananas cost $2.70. Find the price of each.

Problem 5 Solution

Define variables: Let \(a\) = price of one apple (in dollars) and \(b\) = price of one banana (in dollars).

Set up the system:

\[2a + 4b = 2.00\] \[3a + 5b = 2.70\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c} 2 & 4 & | & 2.00 \\ 3 & 5 & | & 2.70 \end{array}\right]\]

Step 2: Make the pivot in row 1, column 1 equal to 1. Divide row 1 by 2.

\[\left[\begin{array}{c c c c} 1 & 2 & | & 1.00 \\ 3 & 5 & | & 2.70 \end{array}\right] \quad R_1 \leftarrow \frac{R_1}{2}\]

Step 3: Eliminate the entry in row 2, column 1. Multiply row 1 by \(-3\) and add to row 2.

\[\left[\begin{array}{c c c c} 1 & 2 & | & 1.00 \\ 0 & -1 & | & -0.30 \end{array}\right] \quad R_2 \leftarrow R_2 - 3R_1\]

Step 4: Make the pivot in row 2, column 2 equal to 1. Divide row 2 by \(-1\).

\[\left[\begin{array}{c c c c} 1 & 2 & | & 1.00 \\ 0 & 1 & | & 0.30 \end{array}\right] \quad R_2 \leftarrow \frac{R_2}{-1}\]

Step 5: Eliminate the entry in row 1, column 2. Multiply row 2 by \(-2\) and add to row 1.

\[\left[\begin{array}{c c c c} 1 & 0 & | & 0.40 \\ 0 & 1 & | & 0.30 \end{array}\right] \quad R_1 \leftarrow R_1 - 2R_2\]

Answer: Each apple costs $0.40 and each banana costs $0.30.

Verification: \(2(0.40) + 4(0.30) = 0.80 + 1.20 = 2.00\) ✓ and \(3(0.40) + 5(0.30) = 1.20 + 1.50 = 2.70\) ✓

6. A bowl of corn flakes, a cup of milk, and an egg provide 16 grams of protein. A cup of milk and two eggs provide 21 grams of protein. Two bowls of corn flakes with two cups of milk provide 16 grams of protein. How much protein is provided by one unit of food?

Problem 6 Solution

Define variables: Let \(c\) = grams of protein in one bowl of corn flakes, \(m\) = grams of protein in one cup of milk, and \(e\) = grams of protein in one egg.

Set up the system:

\[c + m + e = 16\] \[m + 2e = 21\] \[2c + 2m = 16\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c c} 1 & 1 & 1 & | & 16 \\ 0 & 1 & 2 & | & 21 \\ 2 & 2 & 0 & | & 16 \end{array}\right]\]

Step 2: Eliminate the entry in row 3, column 1. Multiply row 1 by \(-2\) and add to row 3.

\[\left[\begin{array}{c c c c c} 1 & 1 & 1 & | & 16 \\ 0 & 1 & 2 & | & 21 \\ 0 & 0 & -2 & | & -16 \end{array}\right] \quad R_3 \leftarrow R_3 - 2R_1\]

Step 3: Eliminate the entry in row 1, column 2. Subtract row 2 from row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & -1 & | & -5 \\ 0 & 1 & 2 & | & 21 \\ 0 & 0 & -2 & | & -16 \end{array}\right] \quad R_1 \leftarrow R_1 - R_2\]

Step 4: Make the pivot in row 3, column 3 equal to 1. Divide row 3 by \(-2\).

\[\left[\begin{array}{c c c c c} 1 & 0 & -1 & | & -5 \\ 0 & 1 & 2 & | & 21 \\ 0 & 0 & 1 & | & 8 \end{array}\right] \quad R_3 \leftarrow \frac{R_3}{-2}\]

Step 5: Eliminate the entry in row 1, column 3. Add row 3 to row 1.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 2 & | & 21 \\ 0 & 0 & 1 & | & 8 \end{array}\right] \quad R_1 \leftarrow R_1 + R_3\]

Step 6: Eliminate the entry in row 2, column 3. Multiply row 3 by \(-2\) and add to row 2.

\[\left[\begin{array}{c c c c c} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & | & 8 \end{array}\right] \quad R_2 \leftarrow R_2 - 2R_3\]

Answer: One bowl of corn flakes provides 3 grams of protein, one cup of milk provides 5 grams of protein, and one egg provides 8 grams of protein.

Verification: \(3 + 5 + 8 = 16\) ✓, \(5 + 2(8) = 5 + 16 = 21\) ✓, and \(2(3) + 2(5) = 6 + 10 = 16\) ✓

7. \(\begin{cases} x + 2y = 10 \\ y + z = 5 \\ z + w = 3 \\ x + w = 5 \end{cases}\)

Problem 7 Solution

Solve the system:

\[x + 2y = 10\] \[y + z = 5\] \[z + w = 3\] \[x + w = 5\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c c c} 1 & 2 & 0 & 0 & | & 10 \\ 0 & 1 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & 1 & | & 3 \\ 1 & 0 & 0 & 1 & | & 5 \end{array}\right]\]

Step 2: Eliminate the entry in row 4, column 1. Subtract row 1 from row 4.

\[\left[\begin{array}{c c c c c c} 1 & 2 & 0 & 0 & | & 10 \\ 0 & 1 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & -2 & 0 & 1 & | & -5 \end{array}\right] \quad R_4 \leftarrow R_4 - R_1\]

Step 3: Eliminate the entry in row 1, column 2. Multiply row 2 by \(-2\) and add to row 1.

\[\left[\begin{array}{c c c c c c} 1 & 0 & -2 & 0 & | & 0 \\ 0 & 1 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & -2 & 0 & 1 & | & -5 \end{array}\right] \quad R_1 \leftarrow R_1 - 2R_2\]

Step 4: Eliminate the entry in row 4, column 2. Add 2 times row 2 to row 4.

\[\left[\begin{array}{c c c c c c} 1 & 0 & -2 & 0 & | & 0 \\ 0 & 1 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & 0 & 2 & 1 & | & 5 \end{array}\right] \quad R_4 \leftarrow R_4 + 2R_2\]

Step 5: Eliminate the entry in row 1, column 3. Add 2 times row 3 to row 1.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 2 & | & 6 \\ 0 & 1 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & 0 & 2 & 1 & | & 5 \end{array}\right] \quad R_1 \leftarrow R_1 + 2R_3\]

Step 6: Eliminate the entry in row 2, column 3. Subtract row 3 from row 2.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 2 & | & 6 \\ 0 & 1 & 0 & -1 & | & 2 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & 0 & 2 & 1 & | & 5 \end{array}\right] \quad R_2 \leftarrow R_2 - R_3\]

Step 7: Eliminate the entry in row 4, column 3. Multiply row 3 by \(-2\) and add to row 4.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 2 & | & 6 \\ 0 & 1 & 0 & -1 & | & 2 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & 0 & 0 & -1 & | & -1 \end{array}\right] \quad R_4 \leftarrow R_4 - 2R_3\]

Step 8: Make the pivot in row 4, column 4 equal to 1. Divide row 4 by \(-1\).

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 2 & | & 6 \\ 0 & 1 & 0 & -1 & | & 2 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & 0 & 0 & 1 & | & 1 \end{array}\right] \quad R_4 \leftarrow \frac{R_4}{-1}\]

Step 9: Eliminate the entry in row 1, column 4. Multiply row 4 by \(-2\) and add to row 1.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 0 & | & 4 \\ 0 & 1 & 0 & -1 & | & 2 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & 0 & 0 & 1 & | & 1 \end{array}\right] \quad R_1 \leftarrow R_1 - 2R_4\]

Step 10: Eliminate the entry in row 2, column 4. Add row 4 to row 2.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 0 & | & 4 \\ 0 & 1 & 0 & 0 & | & 3 \\ 0 & 0 & 1 & 1 & | & 3 \\ 0 & 0 & 0 & 1 & | & 1 \end{array}\right] \quad R_2 \leftarrow R_2 + R_4\]

Step 11: Eliminate the entry in row 3, column 4. Subtract row 4 from row 3.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 0 & | & 4 \\ 0 & 1 & 0 & 0 & | & 3 \\ 0 & 0 & 1 & 0 & | & 2 \\ 0 & 0 & 0 & 1 & | & 1 \end{array}\right] \quad R_3 \leftarrow R_3 - R_4\]

Answer: \(x = 4\), \(y = 3\), \(z = 2\), \(w = 1\)

Verification: \(4 + 2(3) = 10\) ✓, \(3 + 2 = 5\) ✓, \(2 + 1 = 3\) ✓, and \(4 + 1 = 5\) ✓

8. \(\begin{cases} x + w = 6 \\ 2x + y + w = 16 \\ x - 2z = 0 \\ z + w = 5 \end{cases}\)

Problem 8 Solution

Solve the system:

\[x + w = 6\] \[2x + y + w = 16\] \[x - 2z = 0\] \[z + w = 5\]

Step 1: Write the augmented matrix.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 1 & | & 6 \\ 2 & 1 & 0 & 1 & | & 16 \\ 1 & 0 & -2 & 0 & | & 0 \\ 0 & 0 & 1 & 1 & | & 5 \end{array}\right]\]

Step 2: Eliminate the entry in row 2, column 1. Multiply row 1 by \(-2\) and add to row 2.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 1 & | & 6 \\ 0 & 1 & 0 & -1 & | & 4 \\ 1 & 0 & -2 & 0 & | & 0 \\ 0 & 0 & 1 & 1 & | & 5 \end{array}\right] \quad R_2 \leftarrow R_2 - 2R_1\]

Step 3: Eliminate the entry in row 3, column 1. Subtract row 1 from row 3.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 1 & | & 6 \\ 0 & 1 & 0 & -1 & | & 4 \\ 0 & 0 & -2 & -1 & | & -6 \\ 0 & 0 & 1 & 1 & | & 5 \end{array}\right] \quad R_3 \leftarrow R_3 - R_1\]

Step 4: Make the pivot in row 3, column 3 equal to 1. Divide row 3 by \(-2\).

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 1 & | & 6 \\ 0 & 1 & 0 & -1 & | & 4 \\ 0 & 0 & 1 & \tfrac{1}{2} & | & 3 \\ 0 & 0 & 1 & 1 & | & 5 \end{array}\right] \quad R_3 \leftarrow \frac{R_3}{-2}\]

Step 5: Eliminate the entry in row 4, column 3. Subtract row 3 from row 4.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 1 & | & 6 \\ 0 & 1 & 0 & -1 & | & 4 \\ 0 & 0 & 1 & \tfrac{1}{2} & | & 3 \\ 0 & 0 & 0 & \tfrac{1}{2} & | & 2 \end{array}\right] \quad R_4 \leftarrow R_4 - R_3\]

Step 6: Make the pivot in row 4, column 4 equal to 1. Multiply row 4 by 2.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 1 & | & 6 \\ 0 & 1 & 0 & -1 & | & 4 \\ 0 & 0 & 1 & \tfrac{1}{2} & | & 3 \\ 0 & 0 & 0 & 1 & | & 4 \end{array}\right] \quad R_4 \leftarrow 2R_4\]

Step 7: Eliminate the entry in row 1, column 4. Subtract row 4 from row 1.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & -1 & | & 4 \\ 0 & 0 & 1 & \tfrac{1}{2} & | & 3 \\ 0 & 0 & 0 & 1 & | & 4 \end{array}\right] \quad R_1 \leftarrow R_1 - R_4\]

Step 8: Eliminate the entry in row 2, column 4. Add row 4 to row 2.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & 0 & | & 8 \\ 0 & 0 & 1 & \tfrac{1}{2} & | & 3 \\ 0 & 0 & 0 & 1 & | & 4 \end{array}\right] \quad R_2 \leftarrow R_2 + R_4\]

Step 9: Eliminate the entry in row 3, column 4. Subtract \(\tfrac{1}{2}\) times row 4 from row 3.

\[\left[\begin{array}{c c c c c c} 1 & 0 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & 0 & | & 8 \\ 0 & 0 & 1 & 0 & | & 1 \\ 0 & 0 & 0 & 1 & | & 4 \end{array}\right] \quad R_3 \leftarrow R_3 - \tfrac{1}{2}R_4\]

Answer: \(x = 2\), \(y = 8\), \(z = 1\), \(w = 4\)

Verification: \(2 + 4 = 6\) ✓, \(2(2) + 8 + 4 = 16\) ✓, \(2 - 2(1) = 0\) ✓, and \(1 + 4 = 5\) ✓