Finite Math Examples

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

Set up the determinant by breaking it into smaller components.

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

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

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

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

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $2$ .

$0 - 1 (6 - 1 \cdot 3) + 1 | \begin{array}{c} 3 & 3 \\ 4 & 0 \end{array} |$

Multiply $- 1$ by $3$ .

$0 - 1 (6 - 3) + 1 | \begin{array}{c} 3 & 3 \\ 4 & 0 \end{array} |$

Simplify the expression.

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Subtract $3$ from $6$ .

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

Multiply $- 1$ by $3$ .

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

The determinant of $1 | \begin{array}{c} 3 & 3 \\ 4 & 0 \end{array} |$ is $- 12$ .

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

$0 - 3 \cdot 3 \cdot 0 - 4 \cdot 3$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $0$ .

$0 - 3 + 0 - 4 \cdot 3$

Multiply $- 4$ by $3$ .

$0 - 3 + 0 - 12$

Subtract $12$ from $0$ .

$0 - 3 - 12$

Subtract $3$ from $0$ .

$- 3 - 12$

Subtract $12$ from $- 3$ .

$- 15$

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