Finite Math Examples

[\begin{array}{c} 3 & 1 & 4 \\ 1 & 2 & 1 \\ 0 & 1 & 0 \end{array}]

$[\begin{array}{c} 3 & 1 & 4 \\ 1 & 2 & 1 \\ 0 & 1 & 0 \end{array}]$

Step 1

Find the determinant.

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Step 1.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in row

3

$3$ by its cofactor and add.

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Step 1.1.1

Consider the corresponding sign chart.

| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 1.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

Step 1.1.3

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

| \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} |

$| \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} |$

Step 1.1.4

Multiply element

a_{31}

$a_{31}$ by its cofactor.

0 | \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} |

$0 | \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} |$

Step 1.1.5

The minor for

a_{32}

$a_{32}$ is the determinant with row

3

$3$ and column

2

$2$ deleted.

| \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} |

$| \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} |$

Step 1.1.6

Multiply element

a_{32}

$a_{32}$ by its cofactor.

- 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} |

$- 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} |$

Step 1.1.7

The minor for

a_{33}

$a_{33}$ is the determinant with row

3

$3$ and column

3

$3$ deleted.

| \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |

$| \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |$

Step 1.1.8

Multiply element

a_{33}

$a_{33}$ by its cofactor.

0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |

$0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |$

Step 1.1.9

Add the terms together.

0 | \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |

$0 | \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |$

0 | \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |

$0 | \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |$

Step 1.2

Multiply

0

$0$ by

| \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} |

$| \begin{array}{c} 1 & 4 \\ 2 & 1 \end{array} |$ .

0 - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |

$0 - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |$

Step 1.3

Multiply

0

$0$ by

| \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |

$| \begin{array}{c} 3 & 1 \\ 1 & 2 \end{array} |$ .

0 - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0

$0 - 1 | \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} | + 0$

Step 1.4

Evaluate

| \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} |

$| \begin{array}{c} 3 & 4 \\ 1 & 1 \end{array} |$ .

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Step 1.4.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

0 - 1 (3 \cdot 1 - 1 \cdot 4) + 0

$0 - 1 (3 \cdot 1 - 1 \cdot 4) + 0$

Step 1.4.2

Simplify the determinant.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

Multiply

3

$3$ by

1

$1$ .

0 - 1 (3 - 1 \cdot 4) + 0

$0 - 1 (3 - 1 \cdot 4) + 0$

Step 1.4.2.1.2

Multiply

- 1

$- 1$ by

4

$4$ .

0 - 1 (3 - 4) + 0

$0 - 1 (3 - 4) + 0$

0 - 1 (3 - 4) + 0

$0 - 1 (3 - 4) + 0$

Step 1.4.2.2

Subtract

4

$4$ from

3

$3$ .

0 - 1 \cdot - 1 + 0

$0 - 1 \cdot - 1 + 0$

0 - 1 \cdot - 1 + 0

$0 - 1 \cdot - 1 + 0$

0 - 1 \cdot - 1 + 0

$0 - 1 \cdot - 1 + 0$

Step 1.5

Simplify the determinant.

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Step 1.5.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

0 + 1 + 0

$0 + 1 + 0$

Step 1.5.2

Add

0

$0$ and

1

$1$ .

1 + 0

$1 + 0$

Step 1.5.3

Add

1

$1$ and

0

$0$ .

1

$1$

1

$1$

1

$1$

Step 2

Since the determinant is non-zero, the inverse exists.

Step 3

Set up a

3 \times 6

$3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

[\begin{array}{c} 3 & 1 & 4 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 3 & 1 & 4 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 4.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{3}{3} & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & \frac{0}{3} & \frac{0}{3} \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} \frac{3}{3} & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & \frac{0}{3} & \frac{0}{3} \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.2

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 4.2.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 1 - 1 & 2 - \frac{1}{3} & 1 - \frac{4}{3} & 0 - \frac{1}{3} & 1 - 0 & 0 - 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 1 - 1 & 2 - \frac{1}{3} & 1 - \frac{4}{3} & 0 - \frac{1}{3} & 1 - 0 & 0 - 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.2.2

Simplify

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & \frac{5}{3} & - \frac{1}{3} & - \frac{1}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & \frac{5}{3} & - \frac{1}{3} & - \frac{1}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & \frac{5}{3} & - \frac{1}{3} & - \frac{1}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & \frac{5}{3} & - \frac{1}{3} & - \frac{1}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.3

Multiply each element of

R_{2}

$R_{2}$ by

\frac{3}{5}

$\frac{3}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 4.3.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{3}{5}

$\frac{3}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot \frac{5}{3} & \frac{3}{5} (- \frac{1}{3}) & \frac{3}{5} (- \frac{1}{3}) & \frac{3}{5} \cdot 1 & \frac{3}{5} \cdot 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot \frac{5}{3} & \frac{3}{5} (- \frac{1}{3}) & \frac{3}{5} (- \frac{1}{3}) & \frac{3}{5} \cdot 1 & \frac{3}{5} \cdot 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.3.2

Simplify

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.4

Perform the row operation

R_{3} = R_{3} - R_{2}

$R_{3} = R_{3} - R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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Step 4.4.1

Perform the row operation

R_{3} = R_{3} - R_{2}

$R_{3} = R_{3} - R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 - 0 & 1 - 1 & 0 + \frac{1}{5} & 0 + \frac{1}{5} & 0 - \frac{3}{5} & 1 - 0 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 - 0 & 1 - 1 & 0 + \frac{1}{5} & 0 + \frac{1}{5} & 0 - \frac{3}{5} & 1 - 0 \end{array}]$

Step 4.4.2

Simplify

R_{3}

$R_{3}$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 0 & \frac{1}{5} & \frac{1}{5} & - \frac{3}{5} & 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 0 & \frac{1}{5} & \frac{1}{5} & - \frac{3}{5} & 1 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 0 & \frac{1}{5} & \frac{1}{5} & - \frac{3}{5} & 1 \end{array}]

Step 4.5

Multiply each element of

R_{3}

$R_{3}$ by

5

$5$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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Step 4.5.1

Multiply each element of

R_{3}

$R_{3}$ by

5

$5$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 5 \cdot 0 & 5 \cdot 0 & 5 (\frac{1}{5}) & 5 (\frac{1}{5}) & 5 (- \frac{3}{5}) & 5 \cdot 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 5 \cdot 0 & 5 \cdot 0 & 5 (\frac{1}{5}) & 5 (\frac{1}{5}) & 5 (- \frac{3}{5}) & 5 \cdot 1 \end{array}]$

Step 4.5.2

Simplify

R_{3}

$R_{3}$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{3}{5} & 0 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

Step 4.6

Perform the row operation

R_{2} = R_{2} + \frac{1}{5} R_{3}

$R_{2} = R_{2} + \frac{1}{5} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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Step 4.6.1

Perform the row operation

R_{2} = R_{2} + \frac{1}{5} R_{3}

$R_{2} = R_{2} + \frac{1}{5} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 + \frac{1}{5} \cdot 0 & 1 + \frac{1}{5} \cdot 0 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & \frac{3}{5} + \frac{1}{5} \cdot - 3 & 0 + \frac{1}{5} \cdot 5 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 + \frac{1}{5} \cdot 0 & 1 + \frac{1}{5} \cdot 0 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & \frac{3}{5} + \frac{1}{5} \cdot - 3 & 0 + \frac{1}{5} \cdot 5 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

Step 4.6.2

Simplify

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

Step 4.7

Perform the row operation

R_{1} = R_{1} - \frac{4}{3} R_{3}

$R_{1} = R_{1} - \frac{4}{3} R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

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Step 4.7.1

Perform the row operation

R_{1} = R_{1} - \frac{4}{3} R_{3}

$R_{1} = R_{1} - \frac{4}{3} R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

[\begin{array}{c} 1 - \frac{4}{3} \cdot 0 & \frac{1}{3} - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & \frac{1}{3} - \frac{4}{3} \cdot 1 & 0 - \frac{4}{3} \cdot - 3 & 0 - \frac{4}{3} \cdot 5 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 - \frac{4}{3} \cdot 0 & \frac{1}{3} - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & \frac{1}{3} - \frac{4}{3} \cdot 1 & 0 - \frac{4}{3} \cdot - 3 & 0 - \frac{4}{3} \cdot 5 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

Step 4.7.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & \frac{1}{3} & 0 & - 1 & 4 & - \frac{20}{3} \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & 0 & - 1 & 4 & - \frac{20}{3} \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{3} & 0 & - 1 & 4 & - \frac{20}{3} \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{3} & 0 & - 1 & 4 & - \frac{20}{3} \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

Step 4.8

Perform the row operation

R_{1} = R_{1} - \frac{1}{3} R_{2}

$R_{1} = R_{1} - \frac{1}{3} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Step 4.8.1

Perform the row operation

R_{1} = R_{1} - \frac{1}{3} R_{2}

$R_{1} = R_{1} - \frac{1}{3} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & 0 - \frac{1}{3} \cdot 0 & - 1 - \frac{1}{3} \cdot 0 & 4 - \frac{1}{3} \cdot 0 & - \frac{20}{3} - \frac{1}{3} \cdot 1 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & 0 - \frac{1}{3} \cdot 0 & - 1 - \frac{1}{3} \cdot 0 & 4 - \frac{1}{3} \cdot 0 & - \frac{20}{3} - \frac{1}{3} \cdot 1 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

Step 4.8.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & 0 & 0 & - 1 & 4 & - 7 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & - 1 & 4 & - 7 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

[\begin{array}{c} 1 & 0 & 0 & - 1 & 4 & - 7 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & - 1 & 4 & - 7 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

[\begin{array}{c} 1 & 0 & 0 & - 1 & 4 & - 7 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & - 1 & 4 & - 7 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & - 3 & 5 \end{array}]$

Step 5

The right half of the reduced row echelon form is the inverse.

[\begin{array}{c} - 1 & 4 & - 7 \\ 0 & 0 & 1 \\ 1 & - 3 & 5 \end{array}]

$[\begin{array}{c} - 1 & 4 & - 7 \\ 0 & 0 & 1 \\ 1 & - 3 & 5 \end{array}]$

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