Algebra Examples

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

Step 1

Multiply every element in the $2 n d$ row by its cofactor and add.

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

Step 2

Multiply $0$ by $| \begin{array}{c} 1 & 2 \\ 1 & 2 \end{array} |$ .

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

Step 3

Evaluate $| \begin{array}{c} 2 & 5 \\ 2 & 1 \end{array} |$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $1$ .

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

Multiply $- 2$ by $5$ .

$- 3 (2 - 10) + 1 | \begin{array}{c} 1 & 5 \\ 1 & 1 \end{array} | + 0$

Subtract $10$ from $2$ .

$- 3 \cdot - 8 + 1 | \begin{array}{c} 1 & 5 \\ 1 & 1 \end{array} | + 0$

Step 4

Evaluate $| \begin{array}{c} 1 & 5 \\ 1 & 1 \end{array} |$ .

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

$- 3 \cdot - 8 + 1 (1 \cdot 1 - 1 \cdot 5) + 0$

Simplify the determinant.

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Simplify each term.

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Multiply $1$ by $1$ .

$- 3 \cdot - 8 + 1 (1 - 1 \cdot 5) + 0$

Multiply $- 1$ by $5$ .

$- 3 \cdot - 8 + 1 (1 - 5) + 0$

Subtract $5$ from $1$ .

$- 3 \cdot - 8 + 1 \cdot - 4 + 0$

Step 5

Simplify the determinant.

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Simplify each term.

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Multiply $- 3$ by $- 8$ .

$24 + 1 \cdot - 4 + 0$

Multiply $- 4$ by $1$ .

$24 - 4 + 0$

Subtract $4$ from $24$ .

$20 + 0$

Add $20$ and $0$ .

$20$