Linear Algebra Examples

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

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

Step 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

1

$1$ by its cofactor and add.

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Step 1.1

Consider the corresponding sign chart.

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

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

Step 1.2

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

-

$-$ position on the sign chart.

Step 1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

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

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

Step 1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

Step 1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

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

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

Step 1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

Step 1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

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

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

Step 1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

Step 1.9

Add the terms together.

2 ∣ ∣ ∣ \begin{matrix} 0 & - 3 - 1 & 2 \end{matrix} ∣ ∣ ∣ - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣

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

2 ∣ ∣ ∣ \begin{matrix} 0 & - 3 - 1 & 2 \end{matrix} ∣ ∣ ∣ - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣

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

Step 2

Multiply

0

$0$ by

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

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

2 ∣ ∣ ∣ \begin{matrix} 0 & - 3 - 1 & 2 \end{matrix} ∣ ∣ ∣ - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

Step 3

Evaluate

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

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

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Step 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 (0 \cdot 2 - - - 3) - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

Step 3.2

Simplify the determinant.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply

0

$0$ by

2

$2$ .

2 (0 - - - 3) - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

Step 3.2.1.2

Multiply

- - - 3

$- - - 3$ .

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Step 3.2.1.2.1

Multiply

- 1

$- 1$ by

- 3

$- 3$ .

2 (0 - 1 \cdot 3) - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

Step 3.2.1.2.2

Multiply

- 1

$- 1$ by

3

$3$ .

2 (0 - 3) - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

2 (0 - 3) - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

2 (0 - 3) - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

Step 3.2.2

Subtract

3

$3$ from

0

$0$ .

2 \cdot - 3 - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

2 \cdot - 3 - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

2 \cdot - 3 - 5 ∣ ∣ ∣ \begin{matrix} 1 & - 3 2 & 2 \end{matrix} ∣ ∣ ∣ + 0

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

Step 4

Evaluate

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

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

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Step 4.1

The determinant of a

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

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

$2 \cdot - 3 - 5 (1 \cdot 2 - 2 \cdot - 3) + 0$

Step 4.2

Simplify the determinant.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply

$2$ by

$1$ .

$2 \cdot - 3 - 5 (2 - 2 \cdot - 3) + 0$

Step 4.2.1.2

Multiply

$- 2$ by

$- 3$ .

$2 \cdot - 3 - 5 (2 + 6) + 0$

Step 4.2.2

Add

$2$ and

$6$ .

$2 \cdot - 3 - 5 \cdot 8 + 0$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply

$2$ by

$- 3$ .

$- 6 - 5 \cdot 8 + 0$

Step 5.1.2

Multiply

$- 5$ by

$8$ .

$- 6 - 40 + 0$

Step 5.2

Subtract

$40$ from

$- 6$ .

$- 46 + 0$

Step 5.3

Add

$- 46$ and

$0$ .

$- 46$