Finite Math Examples

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

The determinant of $[\begin{array}{c} 0 & 3 & 0 & 1 \\ 4 & 3 & 0 & 3 \\ 1 & 2 & 2 & 4 \\ 1 & 2 & 3 & 4 \end{array}]$ is $34$ .

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Set up the determinant by breaking it into smaller components.

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

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

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

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

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

The determinant of $2 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$ is $- 68$ .

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Set up the determinant by breaking it into smaller components.

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

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

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

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $4$ to get $12$ .

$0 + 2 (0 - 4 (12 - 2 \cdot 1) + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Multiply $- 2$ by $1$ to get $- 2$ .

$0 + 2 (0 - 4 (12 - 2) + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Simplify the expression.

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Subtract $2$ from $12$ to get $10$ .

$0 + 2 (0 - 4 \cdot 10 + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Multiply $- 4$ by $10$ to get $- 40$ .

$0 + 2 (0 - 40 + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

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

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

$0 + 2 (0 - 40 \cdot 3 \cdot 3 - 3 \cdot 1) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $3$ to get $9$ .

$0 + 2 (0 - 40 + 9 - 3 \cdot 1) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Multiply $- 3$ by $1$ to get $- 3$ .

$0 + 2 (0 - 40 + 9 - 3) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Subtract $3$ from $9$ to get $6$ .

$0 + 2 (0 - 40 + 6) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Subtract $40$ from $0$ to get $- 40$ .

$0 + 2 (- 40 + 6) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Add $- 40$ and $6$ to get $- 34$ .

$0 + 2 (- 34) - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

Multiply $2$ by $- 34$ to get $- 68$ .

$0 + 0 - 68 - 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$

The determinant of $- 3 | \begin{array}{c} 0 & 3 & 1 \\ 4 & 3 & 3 \\ 1 & 2 & 4 \end{array} |$ is $102$ .

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Set up the determinant by breaking it into smaller components.

$0 + 0 - 68 - 3 (0 | \begin{array}{c} 3 & 3 \\ 2 & 4 \end{array} | - 4 | \begin{array}{c} 3 & 1 \\ 2 & 4 \end{array} | + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |)$

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

$0 + 0 - 68 - 3 (0 - 4 | \begin{array}{c} 3 & 1 \\ 2 & 4 \end{array} | + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |)$

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

$0 + 0 - 68 - 3 (0 - 4 ((3) (4) - 2 \cdot 1) + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $4$ to get $12$ .

$0 + 0 - 68 - 3 (0 - 4 (12 - 2 \cdot 1) + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |)$

Multiply $- 2$ by $1$ to get $- 2$ .

$0 + 0 - 68 - 3 (0 - 4 (12 - 2) + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |)$

Simplify the expression.

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Subtract $2$ from $12$ to get $10$ .

$0 + 0 - 68 - 3 (0 - 4 \cdot 10 + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |)$

Multiply $- 4$ by $10$ to get $- 40$ .

$0 + 0 - 68 - 3 (0 - 40 + 1 | \begin{array}{c} 3 & 1 \\ 3 & 3 \end{array} |)$

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

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

$0 + 0 - 68 - 3 (0 - 40 \cdot 3 \cdot 3 - 3 \cdot 1)$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $3$ to get $9$ .

$0 + 0 - 68 - 3 (0 - 40 + 9 - 3 \cdot 1)$

Multiply $- 3$ by $1$ to get $- 3$ .

$0 + 0 - 68 - 3 (0 - 40 + 9 - 3)$

Subtract $3$ from $9$ to get $6$ .

$0 + 0 - 68 - 3 (0 - 40 + 6)$

Subtract $40$ from $0$ to get $- 40$ .

$0 + 0 - 68 - 3 (- 40 + 6)$

Add $- 40$ and $6$ to get $- 34$ .

$0 + 0 - 68 - 3 \cdot - 34$

Multiply $- 3$ by $- 34$ to get $102$ .

$0 + 0 - 68 + 102$

Simplify by adding zeros.

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Add $0$ and $0$ to get $0$ .

$0 - 68 + 102$

Subtract $68$ from $0$ to get $- 68$ .

$- 68 + 102$

Add $- 68$ and $102$ to get $34$ .

$34$

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