Finite Math Examples

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

Step 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 column $2$ by its cofactor and add.

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Step 1.1

Consider the corresponding sign chart.

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

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 $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

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

Step 1.4

Multiply element $a_{12}$ by its cofactor.

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

Step 1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Step 1.6

Multiply element $a_{22}$ by its cofactor.

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

Step 1.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 1.8

Multiply element $a_{32}$ by its cofactor.

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

Step 1.9

Add the terms together.

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

Step 2

Multiply $0$ by $| \begin{array}{c} 1 & 1 \\ 4 & 4 \end{array} |$ .

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

Step 3

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

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Step 3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

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 $4$ by $1$ .

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

Step 3.2.1.2

Multiply $- 1$ by $4$ .

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

Step 3.2.2

Subtract $4$ from $4$ .

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

Step 4

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

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Step 4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

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 $1$ by $1$ .

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

Step 4.2.1.2

Multiply $- 1$ by $1$ .

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

Step 4.2.2

Subtract $1$ from $1$ .

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

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $- 1$ by $0$ .

$0 + 3 \cdot 0 + 0$

Step 5.1.2

Multiply $3$ by $0$ .

$0 + 0 + 0$

Step 5.2

Add $0$ and $0$ .

$0 + 0$

Step 5.3

Add $0$ and $0$ .

$0$

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