Finite Math Examples

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

Set up the determinant by breaking it into smaller components.

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

The determinant of $1 | \begin{array}{c} 4 & 2 \\ 1 & 1 \end{array} |$ is $2$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $4$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $4$ .

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

The determinant of $- 3 | \begin{array}{c} 3 & 1 \\ 1 & 1 \end{array} |$ is $- 6$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify the expression.

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Subtract $1$ from $3$ .

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

Multiply $- 3$ by $2$ .

$2 - 6 + 1 | \begin{array}{c} 3 & 1 \\ 4 & 2 \end{array} |$

The determinant of $1 | \begin{array}{c} 3 & 1 \\ 4 & 2 \end{array} |$ is $2$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $2$ .

$2 - 6 + 6 - 4 \cdot 1$

Multiply $- 4$ by $1$ .

$2 - 6 + 6 - 4$

Subtract $4$ from $6$ .

$2 - 6 + 2$

Subtract $6$ from $2$ .

$- 4 + 2$

Add $- 4$ and $2$ .

$- 2$

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