Finite Math Examples

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

Set up the determinant by breaking it into smaller components.

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

Since the matrix is multiplied by $0$ , the determinant is $0$ .

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

The determinant of $4 | \begin{array}{c} 3 & 1 \\ 4 & 3 \end{array} |$ is $20$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $3$ .

$0 + 4 (9 - 4 \cdot 1) - 4 | \begin{array}{c} 3 & 1 \\ 2 & 4 \end{array} |$

Multiply $- 4$ by $1$ .

$0 + 4 (9 - 4) - 4 | \begin{array}{c} 3 & 1 \\ 2 & 4 \end{array} |$

Simplify the expression.

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Subtract $4$ from $9$ .

$0 + 4 \cdot 5 - 4 | \begin{array}{c} 3 & 1 \\ 2 & 4 \end{array} |$

Multiply $4$ by $5$ .

$0 + 20 - 4 | \begin{array}{c} 3 & 1 \\ 2 & 4 \end{array} |$

The determinant of $- 4 | \begin{array}{c} 3 & 1 \\ 2 & 4 \end{array} |$ is $- 40$ .

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

$0 + 20 - 4 ((3) (4) - 2 \cdot 1)$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $4$ .

$0 + 20 - 4 (12 - 2 \cdot 1)$

Multiply $- 2$ by $1$ .

$0 + 20 - 4 (12 - 2)$

Simplify the expression.

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Subtract $2$ from $12$ .

$0 + 20 - 4 \cdot 10$

Multiply $- 4$ by $10$ .

$0 + 20 - 40$

Add $0$ and $20$ .

$20 - 40$

Subtract $40$ from $20$ .

$- 20$

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