Algebra Examples

$[\begin{array}{c} 1 & - 1 & 3 \\ 2 & 5 & 0 \\ - 3 & 1 & 2 \end{array}]$

Set up the determinant by breaking it into smaller components.

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

Find the determinant of $3 | \begin{array}{c} 2 & 5 \\ - 3 & 1 \end{array} |$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $1$ .

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

Multiply $3$ by $5$ .

$3 (2 + 15) + 0 | \begin{array}{c} 1 & - 1 \\ - 3 & 1 \end{array} | + 2 | \begin{array}{c} 1 & - 1 \\ 2 & 5 \end{array} |$

Simplify the expression.

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Add $2$ and $15$ .

$3 \cdot 17 + 0 | \begin{array}{c} 1 & - 1 \\ - 3 & 1 \end{array} | + 2 | \begin{array}{c} 1 & - 1 \\ 2 & 5 \end{array} |$

Multiply $3$ by $17$ .

$51 + 0 | \begin{array}{c} 1 & - 1 \\ - 3 & 1 \end{array} | + 2 | \begin{array}{c} 1 & - 1 \\ 2 & 5 \end{array} |$

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

$51 + 0 + 2 | \begin{array}{c} 1 & - 1 \\ 2 & 5 \end{array} |$

Find the determinant of $2 | \begin{array}{c} 1 & - 1 \\ 2 & 5 \end{array} |$ .

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

$51 + 0 + 2 ((1) (5) - 2 \cdot - 1)$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $1$ .

$51 + 0 + 2 (5 - 2 \cdot - 1)$

Multiply $- 2$ by $- 1$ .

$51 + 0 + 2 (5 + 2)$

Simplify the expression.

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Add $5$ and $2$ .

$51 + 0 + 2 \cdot 7$

Multiply $2$ by $7$ .

$51 + 0 + 14$

Add $51$ and $0$ .

$51 + 14$

Add $51$ and $14$ .

$65$