Examples

⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 4 & 3 & 1 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

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 column

2

$2$ by its cofactor and add.

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Step 1.1.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

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

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_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 4 & 1 2 & 1 \end{matrix} ∣ ∣ ∣

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

Step 1.1.4

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- 2 ∣ ∣ ∣ \begin{matrix} 4 & 1 2 & 1 \end{matrix} ∣ ∣ ∣

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

Step 1.1.5

The minor for

a_{22}

$a_{22}$ is the determinant with row

2

$2$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣

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

Step 1.1.6

Multiply element

a_{22}

$a_{22}$ by its cofactor.

3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣

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

Step 1.1.7

The minor for

a_{32}

$a_{32}$ is the determinant with row

3

$3$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 1 4 & 1 \end{matrix} ∣ ∣ ∣

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

Step 1.1.8

Multiply element

a_{32}

$a_{32}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 2 & 1 4 & 1 \end{matrix} ∣ ∣ ∣

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

Step 1.1.9

Add the terms together.

- 2 ∣ ∣ ∣ \begin{matrix} 4 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 2 & 1 4 & 1 \end{matrix} ∣ ∣ ∣

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

- 2 ∣ ∣ ∣ \begin{matrix} 4 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 2 & 1 4 & 1 \end{matrix} ∣ ∣ ∣

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

Step 1.2

Multiply

0

$0$ by

∣ ∣ ∣ \begin{matrix} 2 & 1 4 & 1 \end{matrix} ∣ ∣ ∣

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

- 2 ∣ ∣ ∣ \begin{matrix} 4 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 0

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

Step 1.3

Evaluate

∣ ∣ ∣ \begin{matrix} 4 & 1 2 & 1 \end{matrix} ∣ ∣ ∣

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

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Step 1.3.1

The determinant of a

2 \times 2

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

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

- 2 (4 \cdot 1 - 2 \cdot 1) + 3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 0

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

Step 1.3.2

Simplify the determinant.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Multiply

4

$4$ by

1

$1$ .

- 2 (4 - 2 \cdot 1) + 3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 0

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

Step 1.3.2.1.2

Multiply

- 2

$- 2$ by

1

$1$ .

- 2 (4 - 2) + 3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 0

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

- 2 (4 - 2) + 3 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 \end{matrix} ∣ ∣ ∣ + 0

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

Step 1.3.2.2

Subtract

$2$ from

$4$ .

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

Step 1.4

Evaluate

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

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Step 1.4.1

The determinant of a

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

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

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

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

$2$ by

$1$ .

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

Step 1.4.2.1.2

Multiply

$- 2$ by

$1$ .

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

Step 1.4.2.2

Subtract

$2$ from

$2$ .

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

Step 1.5

Simplify the determinant.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

Multiply

$- 2$ by

$2$ .

$- 4 + 3 \cdot 0 + 0$

Step 1.5.1.2

Multiply

$3$ by

$0$ .

$- 4 + 0 + 0$

Step 1.5.2

Add

$- 4$ and

$0$ .

$- 4 + 0$

Step 1.5.3

Add

$- 4$ and

$0$ .

$- 4$

Step 2

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

Step 3

Set up a

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

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of

$R_{1}$ by

$\frac{1}{2}$ to make the entry at

$1, 1$ a

$1$ .

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Step 4.1.1

Multiply each element of

$R_{1}$ by

$\frac{1}{2}$ to make the entry at

$1, 1$ a

$1$ .

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

Step 4.1.2

Simplify

$R_{1}$ .

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

Step 4.2

Perform the row operation

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

$2, 1$ a

$0$ .

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Step 4.2.1

Perform the row operation

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

$2, 1$ a

$0$ .

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

Step 4.2.2

Simplify

$R_{2}$ .

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

Step 4.3

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 4.3.1

Perform the row operation

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

$3, 1$ a

$0$ .

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

Step 4.3.2

Simplify

$R_{3}$ .

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

Step 4.4

Multiply each element of

$R_{2}$ by

$- 1$ to make the entry at

$2, 2$ a

$1$ .

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Step 4.4.1

Multiply each element of

$R_{2}$ by

$- 1$ to make the entry at

$2, 2$ a

$1$ .

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

Step 4.4.2

Simplify

$R_{2}$ .

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

Step 4.5

Perform the row operation

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

$3, 2$ a

$0$ .

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Step 4.5.1

Perform the row operation

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

$3, 2$ a

$0$ .

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

Step 4.5.2

Simplify

$R_{3}$ .

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

Step 4.6

Multiply each element of

$R_{3}$ by

$\frac{1}{2}$ to make the entry at

$3, 3$ a

$1$ .

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Step 4.6.1

Multiply each element of

$R_{3}$ by

$\frac{1}{2}$ to make the entry at

$3, 3$ a

$1$ .

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

Step 4.6.2

Simplify

$R_{3}$ .

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

Step 4.7

Perform the row operation

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

$2, 3$ a

$0$ .

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Step 4.7.1

Perform the row operation

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

$2, 3$ a

$0$ .

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

Step 4.7.2

Simplify

$R_{2}$ .

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

Step 4.8

Perform the row operation

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

$1, 3$ a

$0$ .

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Step 4.8.1

Perform the row operation

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

$1, 3$ a

$0$ .

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

Step 4.8.2

Simplify

$R_{1}$ .

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

Step 4.9

Perform the row operation

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

$1, 2$ a

$0$ .

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Step 4.9.1

Perform the row operation

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

$1, 2$ a

$0$ .

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

Step 4.9.2

Simplify

$R_{1}$ .

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

Step 5

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

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

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