Linear Algebra Examples

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 0 & - 2 3 & - 12 & - 2 & - 6 - 2 & 10 & 2 & 5 - 1 & 6 & 1 & 3 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 0 & - 2 \\ 3 & - 12 & - 2 & - 6 \\ - 2 & 10 & 2 & 5 \\ - 1 & 6 & 1 & 3 \end{array}]$

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

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

$1$ by its cofactor and add.

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Step 1.1.1

Consider the corresponding sign chart.

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

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

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} |$

Step 1.1.4

Multiply element

$a_{11}$ by its cofactor.

$1 | \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} |$

Step 1.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} |$

Step 1.1.6

Multiply element

$a_{12}$ by its cofactor.

$3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} |$

Step 1.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} |$

Step 1.1.8

Multiply element

$a_{13}$ by its cofactor.

$0 | \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} |$

Step 1.1.9

The minor for

$a_{14}$ is the determinant with row

$1$ and column

$4$ deleted.

$| \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.1.10

Multiply element

$a_{14}$ by its cofactor.

$2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.1.11

Add the terms together.

$1 | \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} | + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 | \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} | + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.2

Multiply

$0$ by

$| \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} |$ .

$1 | \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} | + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3

Evaluate

$| \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} |$ .

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Step 1.3.1

Choose the row or column with the most

$0$ elements. If there are no

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

$1$ by its cofactor and add.

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Step 1.3.1.1

Consider the corresponding sign chart.

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

Step 1.3.1.2

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

$-$ position on the sign chart.

Step 1.3.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

Step 1.3.1.4

Multiply element

$a_{11}$ by its cofactor.

$- 12 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

Step 1.3.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} |$

Step 1.3.1.6

Multiply element

$a_{12}$ by its cofactor.

$2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} |$

Step 1.3.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

Step 1.3.1.8

Multiply element

$a_{13}$ by its cofactor.

$- 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

Step 1.3.1.9

Add the terms together.

$1 (- 12 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} | + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.2

Evaluate

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$ .

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Step 1.3.2.1

The determinant of a

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

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

$1 (- 12 (2 \cdot 3 - 1 \cdot 5) + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.2.2

Simplify the determinant.

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Step 1.3.2.2.1

Simplify each term.

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Step 1.3.2.2.1.1

Multiply

$2$ by

$3$ .

$1 (- 12 (6 - 1 \cdot 5) + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.2.2.1.2

Multiply

$- 1$ by

$5$ .

$1 (- 12 (6 - 5) + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.2.2.2

Subtract

$5$ from

$6$ .

$1 (- 12 \cdot 1 + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.3

Evaluate

$| \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} |$ .

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Step 1.3.3.1

The determinant of a

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

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

$1 (- 12 \cdot 1 + 2 (10 \cdot 3 - 6 \cdot 5) - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.3.2

Simplify the determinant.

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Step 1.3.3.2.1

Simplify each term.

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Step 1.3.3.2.1.1

Multiply

$10$ by

$3$ .

$1 (- 12 \cdot 1 + 2 (30 - 6 \cdot 5) - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.3.2.1.2

Multiply

$- 6$ by

$5$ .

$1 (- 12 \cdot 1 + 2 (30 - 30) - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.3.2.2

Subtract

$30$ from

$30$ .

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.4

Evaluate

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$ .

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Step 1.3.4.1

The determinant of a

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

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

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 (10 \cdot 1 - 6 \cdot 2)) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.4.2

Simplify the determinant.

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Step 1.3.4.2.1

Simplify each term.

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Step 1.3.4.2.1.1

Multiply

$10$ by

$1$ .

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 (10 - 6 \cdot 2)) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.4.2.1.2

Multiply

$- 6$ by

$2$ .

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 (10 - 12)) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.4.2.2

Subtract

$12$ from

$10$ .

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 \cdot - 2) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.5

Simplify the determinant.

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Step 1.3.5.1

Simplify each term.

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Step 1.3.5.1.1

Multiply

$- 12$ by

$1$ .

$1 (- 12 + 2 \cdot 0 - 6 \cdot - 2) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.5.1.2

Multiply

$2$ by

$0$ .

$1 (- 12 + 0 - 6 \cdot - 2) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.5.1.3

Multiply

$- 6$ by

$- 2$ .

$1 (- 12 + 0 + 12) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.5.2

Add

$- 12$ and

$0$ .

$1 (- 12 + 12) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.3.5.3

Add

$- 12$ and

$12$ .

$1 \cdot 0 + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4

Evaluate

$| \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} |$ .

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Step 1.4.1

Choose the row or column with the most

$0$ elements. If there are no

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

$1$ by its cofactor and add.

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Step 1.4.1.1

Consider the corresponding sign chart.

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

Step 1.4.1.2

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

$-$ position on the sign chart.

Step 1.4.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

Step 1.4.1.4

Multiply element

$a_{11}$ by its cofactor.

$3 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

Step 1.4.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} |$

Step 1.4.1.6

Multiply element

$a_{12}$ by its cofactor.

$2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} |$

Step 1.4.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

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

Step 1.4.1.8

Multiply element

$a_{13}$ by its cofactor.

$- 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |$

Step 1.4.1.9

Add the terms together.

$1 \cdot 0 + 3 (3 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} | + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.2

Evaluate

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$ .

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Step 1.4.2.1

The determinant of a

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

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

$1 \cdot 0 + 3 (3 (2 \cdot 3 - 1 \cdot 5) + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.2.2

Simplify the determinant.

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Step 1.4.2.2.1

Simplify each term.

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Step 1.4.2.2.1.1

Multiply

$2$ by

$3$ .

$1 \cdot 0 + 3 (3 (6 - 1 \cdot 5) + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.2.2.1.2

Multiply

$- 1$ by

$5$ .

$1 \cdot 0 + 3 (3 (6 - 5) + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.2.2.2

Subtract

$5$ from

$6$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.3

Evaluate

$| \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} |$ .

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Step 1.4.3.1

The determinant of a

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

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

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 2 \cdot 3 - (- 1 \cdot 5)) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.3.2

Simplify the determinant.

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Step 1.4.3.2.1

Simplify each term.

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Step 1.4.3.2.1.1

Multiply

$- 2$ by

$3$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 6 - (- 1 \cdot 5)) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.3.2.1.2

Multiply

$- (- 1 \cdot 5)$ .

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Step 1.4.3.2.1.2.1

Multiply

$- 1$ by

$5$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 6 - - 5) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.3.2.1.2.2

Multiply

$- 1$ by

$- 5$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 6 + 5) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.3.2.2

Add

$- 6$ and

$5$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.4

Evaluate

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

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Step 1.4.4.1

The determinant of a

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

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

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 \cdot 1 - (- 1 \cdot 2))) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.4.2

Simplify the determinant.

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Step 1.4.4.2.1

Simplify each term.

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Step 1.4.4.2.1.1

Multiply

$- 2$ by

$1$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 - (- 1 \cdot 2))) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.4.2.1.2

Multiply

$- (- 1 \cdot 2)$ .

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Step 1.4.4.2.1.2.1

Multiply

$- 1$ by

$2$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 - - 2)) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.4.2.1.2.2

Multiply

$- 1$ by

$- 2$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 + 2)) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.4.2.2

Add

$- 2$ and

$2$ .

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 \cdot 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.5

Simplify the determinant.

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Step 1.4.5.1

Simplify each term.

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Step 1.4.5.1.1

Multiply

$3$ by

$1$ .

$1 \cdot 0 + 3 (3 + 2 \cdot - 1 - 6 \cdot 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.5.1.2

Multiply

$2$ by

$- 1$ .

$1 \cdot 0 + 3 (3 - 2 - 6 \cdot 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.5.1.3

Multiply

$- 6$ by

$0$ .

$1 \cdot 0 + 3 (3 - 2 + 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.5.2

Subtract

$2$ from

$3$ .

$1 \cdot 0 + 3 (1 + 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.4.5.3

Add

$1$ and

$0$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

Step 1.5

Evaluate

$| \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$ .

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Step 1.5.1

Choose the row or column with the most

$0$ elements. If there are no

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

$1$ by its cofactor and add.

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Step 1.5.1.1

Consider the corresponding sign chart.

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

Step 1.5.1.2

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

$-$ position on the sign chart.

Step 1.5.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

Step 1.5.1.4

Multiply element

$a_{11}$ by its cofactor.

$3 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

Step 1.5.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

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

Step 1.5.1.6

Multiply element

$a_{12}$ by its cofactor.

$12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |$

Step 1.5.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |$

Step 1.5.1.8

Multiply element

$a_{13}$ by its cofactor.

$- 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |$

Step 1.5.1.9

Add the terms together.

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} | + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.2

Evaluate

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$ .

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Step 1.5.2.1

The determinant of a

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

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

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 (10 \cdot 1 - 6 \cdot 2) + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.2.2

Simplify the determinant.

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Step 1.5.2.2.1

Simplify each term.

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Step 1.5.2.2.1.1

Multiply

$10$ by

$1$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 (10 - 6 \cdot 2) + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.2.2.1.2

Multiply

$- 6$ by

$2$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 (10 - 12) + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.2.2.2

Subtract

$12$ from

$10$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.3

Evaluate

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

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Step 1.5.3.1

The determinant of a

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

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

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 \cdot 1 - (- 1 \cdot 2)) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.3.2

Simplify the determinant.

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Step 1.5.3.2.1

Simplify each term.

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Step 1.5.3.2.1.1

Multiply

$- 2$ by

$1$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 - (- 1 \cdot 2)) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.3.2.1.2

Multiply

$- (- 1 \cdot 2)$ .

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Step 1.5.3.2.1.2.1

Multiply

$- 1$ by

$2$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 - - 2) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.3.2.1.2.2

Multiply

$- 1$ by

$- 2$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 + 2) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.3.2.2

Add

$- 2$ and

$2$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

Step 1.5.4

Evaluate

$| \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |$ .

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Step 1.5.4.1

The determinant of a

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

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

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 2 \cdot 6 - (- 1 \cdot 10)))$

Step 1.5.4.2

Simplify the determinant.

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Step 1.5.4.2.1

Simplify each term.

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Step 1.5.4.2.1.1

Multiply

$- 2$ by

$6$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 12 - (- 1 \cdot 10)))$

Step 1.5.4.2.1.2

Multiply

$- (- 1 \cdot 10)$ .

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Step 1.5.4.2.1.2.1

Multiply

$- 1$ by

$10$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 12 - - 10))$

Step 1.5.4.2.1.2.2

Multiply

$- 1$ by

$- 10$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 12 + 10))$

Step 1.5.4.2.2

Add

$- 12$ and

$10$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 \cdot - 2)$

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Simplify each term.

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Step 1.5.5.1.1

Multiply

$3$ by

$- 2$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 12 \cdot 0 - 2 \cdot - 2)$

Step 1.5.5.1.2

Multiply

$12$ by

$0$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 0 - 2 \cdot - 2)$

Step 1.5.5.1.3

Multiply

$- 2$ by

$- 2$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 0 + 4)$

Step 1.5.5.2

Add

$- 6$ and

$0$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 4)$

Step 1.5.5.3

Add

$- 6$ and

$4$ .

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 \cdot - 2$

Step 1.6

Simplify the determinant.

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Step 1.6.1

Simplify each term.

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Step 1.6.1.1

Multiply

$0$ by

$1$ .

$0 + 3 \cdot 1 + 0 + 2 \cdot - 2$

Step 1.6.1.2

Multiply

$3$ by

$1$ .

$0 + 3 + 0 + 2 \cdot - 2$

Step 1.6.1.3

Multiply

$2$ by

$- 2$ .

$0 + 3 + 0 - 4$

Step 1.6.2

Add

$0$ and

$3$ .

$3 + 0 - 4$

Step 1.6.3

Add

$3$ and

$0$ .

$3 - 4$

Step 1.6.4

Subtract

$4$ from

$3$ .

$- 1$

Step 2

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

Step 3

Set up a

$4 \times 8$ matrix where the left half is the original matrix and the right half is its identity matrix.

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 3 & - 12 & - 2 & - 6 & 0 & 1 & 0 & 0 \\ - 2 & 10 & 2 & 5 & 0 & 0 & 1 & 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Perform the row operation

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

$2, 1$ a

$0$ .

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Step 4.1.1

Perform the row operation

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

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 3 - 3 \cdot 1 & - 12 - 3 \cdot - 3 & - 2 - 3 \cdot 0 & - 6 - 3 \cdot - 2 & 0 - 3 \cdot 1 & 1 - 3 \cdot 0 & 0 - 3 \cdot 0 & 0 - 3 \cdot 0 \\ - 2 & 10 & 2 & 5 & 0 & 0 & 1 & 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.1.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 0 & - 3 & - 2 & 0 & - 3 & 1 & 0 & 0 \\ - 2 & 10 & 2 & 5 & 0 & 0 & 1 & 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.2

Perform the row operation

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

$3, 1$ a

$0$ .

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Step 4.2.1

Perform the row operation

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

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 0 & - 3 & - 2 & 0 & - 3 & 1 & 0 & 0 \\ - 2 + 2 \cdot 1 & 10 + 2 \cdot - 3 & 2 + 2 \cdot 0 & 5 + 2 \cdot - 2 & 0 + 2 \cdot 1 & 0 + 2 \cdot 0 & 1 + 2 \cdot 0 & 0 + 2 \cdot 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.2.2

Simplify

$R_{3}$ .

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

Step 4.3

Perform the row operation

$R_{4} = R_{4} + R_{1}$ to make the entry at

$4, 1$ a

$0$ .

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Step 4.3.1

Perform the row operation

$R_{4} = R_{4} + R_{1}$ to make the entry at

$4, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 0 & - 3 & - 2 & 0 & - 3 & 1 & 0 & 0 \\ 0 & 4 & 2 & 1 & 2 & 0 & 1 & 0 \\ - 1 + 1 \cdot 1 & 6 - 3 & 1 + 0 & 3 - 2 & 0 + 1 \cdot 1 & 0 + 0 & 0 + 0 & 1 + 0 \end{array}]$

Step 4.3.2

Simplify

$R_{4}$ .

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

Step 4.4

Multiply each element of

$R_{2}$ by

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

$2, 2$ a

$1$ .

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Step 4.4.1

Multiply each element of

$R_{2}$ by

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

$2, 2$ a

$1$ .

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

Step 4.4.2

Simplify

$R_{2}$ .

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

Step 4.5

Perform the row operation

$R_{3} = R_{3} - 4 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Step 4.5.1

Perform the row operation

$R_{3} = R_{3} - 4 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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

Step 4.5.2

Simplify

$R_{3}$ .

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

Step 4.6

Perform the row operation

$R_{4} = R_{4} - 3 R_{2}$ to make the entry at

$4, 2$ a

$0$ .

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Step 4.6.1

Perform the row operation

$R_{4} = R_{4} - 3 R_{2}$ to make the entry at

$4, 2$ a

$0$ .

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

Step 4.6.2

Simplify

$R_{4}$ .

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

Step 4.7

Multiply each element of

$R_{3}$ by

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

$3, 3$ a

$1$ .

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Step 4.7.1

Multiply each element of

$R_{3}$ by

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

$3, 3$ a

$1$ .

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

Step 4.7.2

Simplify

$R_{3}$ .

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

Step 4.8

Perform the row operation

$R_{4} = R_{4} + R_{3}$ to make the entry at

$4, 3$ a

$0$ .

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Step 4.8.1

Perform the row operation

$R_{4} = R_{4} + R_{3}$ to make the entry at

$4, 3$ a

$0$ .

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

Step 4.8.2

Simplify

$R_{4}$ .

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

Step 4.9

Multiply each element of

$R_{4}$ by

$- 2$ to make the entry at

$4, 4$ a

$1$ .

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Step 4.9.1

Multiply each element of

$R_{4}$ by

$- 2$ to make the entry at

$4, 4$ a

$1$ .

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

Step 4.9.2

Simplify

$R_{4}$ .

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

Step 4.10

Perform the row operation

$R_{3} = R_{3} + \frac{3}{2} R_{4}$ to make the entry at

$3, 4$ a

$0$ .

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Step 4.10.1

Perform the row operation

$R_{3} = R_{3} + \frac{3}{2} R_{4}$ to make the entry at

$3, 4$ a

$0$ .

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

Step 4.10.2

Simplify

$R_{3}$ .

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

Step 4.11

Perform the row operation

$R_{1} = R_{1} + 2 R_{4}$ to make the entry at

$1, 4$ a

$0$ .

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Step 4.11.1

Perform the row operation

$R_{1} = R_{1} + 2 R_{4}$ to make the entry at

$1, 4$ a

$0$ .

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

Step 4.11.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & - 3 & 0 & 0 & - 3 & 4 & 6 & - 4 \\ 0 & 1 & \frac{2}{3} & 0 & 1 & - \frac{1}{3} & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 3 & - 3 \\ 0 & 0 & 0 & 1 & - 2 & 2 & 3 & - 2 \end{array}]$

Step 4.12

Perform the row operation

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

$2, 3$ a

$0$ .

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Step 4.12.1

Perform the row operation

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

$2, 3$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & 0 & - 3 & 4 & 6 & - 4 \\ 0 - \frac{2}{3} \cdot 0 & 1 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & 0 - \frac{2}{3} \cdot 0 & 1 - \frac{2}{3} \cdot 0 & - \frac{1}{3} - \frac{2}{3} \cdot 1 & 0 - \frac{2}{3} \cdot 3 & 0 - \frac{2}{3} \cdot - 3 \\ 0 & 0 & 1 & 0 & 0 & 1 & 3 & - 3 \\ 0 & 0 & 0 & 1 & - 2 & 2 & 3 & - 2 \end{array}]$

Step 4.12.2

Simplify

$R_{2}$ .

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

Step 4.13

Perform the row operation

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

$1, 2$ a

$0$ .

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Step 4.13.1

Perform the row operation

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

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 0 + 3 \cdot 0 & 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 4 + 3 \cdot - 1 & 6 + 3 \cdot - 2 & - 4 + 3 \cdot 2 \\ 0 & 1 & 0 & 0 & 1 & - 1 & - 2 & 2 \\ 0 & 0 & 1 & 0 & 0 & 1 & 3 & - 3 \\ 0 & 0 & 0 & 1 & - 2 & 2 & 3 & - 2 \end{array}]$

Step 4.13.2

Simplify

$R_{1}$ .

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

Step 5

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

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