Linear Algebra Examples

$[\begin{array}{c} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 9 \end{array}]$

Step 1

Find the determinant.

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Step 1.1

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 1.1.1

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.

$| \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} |$

Step 1.1.4

Multiply element

$a_{11}$ by its cofactor.

$1 | \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} |$

Step 1.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} |$

Step 1.1.6

Multiply element

$a_{12}$ by its cofactor.

$- 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} |$

Step 1.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.1.8

Multiply element

$a_{13}$ by its cofactor.

$3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.1.9

Add the terms together.

$1 | \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} | - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.2

Evaluate

$| \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} |$ .

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Step 1.2.1

The determinant of a

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

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

$1 (5 \cdot 9 - 7 \cdot 7) - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.2.2

Simplify the determinant.

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Step 1.2.2.1

Simplify each term.

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Step 1.2.2.1.1

Multiply

$5$ by

$9$ .

$1 (45 - 7 \cdot 7) - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.2.2.1.2

Multiply

$- 7$ by

$7$ .

$1 (45 - 49) - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.2.2.2

Subtract

$49$ from

$45$ .

$1 \cdot - 4 - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.3

Evaluate

$| \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} |$ .

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Step 1.3.1

The determinant of a

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

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

$1 \cdot - 4 - 2 (2 \cdot 9 - 3 \cdot 7) + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

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

$2$ by

$9$ .

$1 \cdot - 4 - 2 (18 - 3 \cdot 7) + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.3.2.1.2

Multiply

$- 3$ by

$7$ .

$1 \cdot - 4 - 2 (18 - 21) + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.3.2.2

Subtract

$21$ from

$18$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Step 1.4

Evaluate

$| \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$ .

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Step 1.4.1

The determinant of a

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

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

$1 \cdot - 4 - 2 \cdot - 3 + 3 (2 \cdot 7 - 3 \cdot 5)$

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

$7$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 (14 - 3 \cdot 5)$

Step 1.4.2.1.2

Multiply

$- 3$ by

$5$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 (14 - 15)$

Step 1.4.2.2

Subtract

$15$ from

$14$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 \cdot - 1$

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

$- 4$ by

$1$ .

$- 4 - 2 \cdot - 3 + 3 \cdot - 1$

Step 1.5.1.2

Multiply

$- 2$ by

$- 3$ .

$- 4 + 6 + 3 \cdot - 1$

Step 1.5.1.3

Multiply

$3$ by

$- 1$ .

$- 4 + 6 - 3$

Step 1.5.2

Add

$- 4$ and

$6$ .

$2 - 3$

Step 1.5.3

Subtract

$3$ from

$2$ .

$- 1$

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.

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 7 & 0 & 1 & 0 \\ 3 & 7 & 9 & 0 & 0 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Perform the row operation

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

$2, 1$ a

$0$ .

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Step 4.1.1

Perform the row operation

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

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 - 2 \cdot 1 & 5 - 2 \cdot 2 & 7 - 2 \cdot 3 & 0 - 2 \cdot 1 & 1 - 2 \cdot 0 & 0 - 2 \cdot 0 \\ 3 & 7 & 9 & 0 & 0 & 1 \end{array}]$

Step 4.1.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & - 2 & 1 & 0 \\ 3 & 7 & 9 & 0 & 0 & 1 \end{array}]$

Step 4.2

Perform the row operation

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

$3, 1$ a

$0$ .

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Step 4.2.1

Perform the row operation

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

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & - 2 & 1 & 0 \\ 3 - 3 \cdot 1 & 7 - 3 \cdot 2 & 9 - 3 \cdot 3 & 0 - 3 \cdot 1 & 0 - 3 \cdot 0 & 1 - 3 \cdot 0 \end{array}]$

Step 4.2.2

Simplify

$R_{3}$ .

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

Step 4.3

Perform the row operation

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

$3, 2$ a

$0$ .

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Step 4.3.1

Perform the row operation

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

$3, 2$ a

$0$ .

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

Step 4.3.2

Simplify

$R_{3}$ .

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

Step 4.4

Multiply each element of

$R_{3}$ by

$- 1$ to make the entry at

$3, 3$ a

$1$ .

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Step 4.4.1

Multiply each element of

$R_{3}$ by

$- 1$ to make the entry at

$3, 3$ a

$1$ .

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

Step 4.4.2

Simplify

$R_{3}$ .

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

Step 4.5

Perform the row operation

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

$2, 3$ a

$0$ .

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Step 4.5.1

Perform the row operation

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

$2, 3$ a

$0$ .

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

Step 4.5.2

Simplify

$R_{2}$ .

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

Step 4.6

Perform the row operation

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

$1, 3$ a

$0$ .

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Step 4.6.1

Perform the row operation

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

$1, 3$ a

$0$ .

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

Step 4.6.2

Simplify

$R_{1}$ .

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

Step 4.7

Perform the row operation

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

$1, 2$ a

$0$ .

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Step 4.7.1

Perform the row operation

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

$1, 2$ a

$0$ .

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

Step 4.7.2

Simplify

$R_{1}$ .

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

Step 5

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

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