Linear Algebra Examples

$[\begin{array}{c} - \frac{3}{17} & \frac{5}{17} \\ \frac{4}{17} & - \frac{1}{17} \end{array}]$

Step 1

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

$- \frac{3}{17} (- \frac{1}{17}) - \frac{4}{17} \cdot \frac{5}{17}$

Step 2

Simplify the determinant.

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Step 2.1

Simplify each term.

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Step 2.1.1

Multiply $- \frac{3}{17} (- \frac{1}{17})$ .

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Step 2.1.1.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{3}{17}) \frac{1}{17} - \frac{4}{17} \cdot \frac{5}{17}$

Step 2.1.1.2

Multiply $\frac{3}{17}$ by $1$ .

$\frac{3}{17} \cdot \frac{1}{17} - \frac{4}{17} \cdot \frac{5}{17}$

Step 2.1.1.3

Multiply $\frac{3}{17}$ by $\frac{1}{17}$ .

$\frac{3}{17 \cdot 17} - \frac{4}{17} \cdot \frac{5}{17}$

Step 2.1.1.4

Multiply $17$ by $17$ .

$\frac{3}{289} - \frac{4}{17} \cdot \frac{5}{17}$

Step 2.1.2

Multiply $- \frac{4}{17} \cdot \frac{5}{17}$ .

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Step 2.1.2.1

Multiply $\frac{5}{17}$ by $\frac{4}{17}$ .

$\frac{3}{289} - \frac{5 \cdot 4}{17 \cdot 17}$

Step 2.1.2.2

Multiply $5$ by $4$ .

$\frac{3}{289} - \frac{20}{17 \cdot 17}$

Step 2.1.2.3

Multiply $17$ by $17$ .

$\frac{3}{289} - \frac{20}{289}$

Step 2.2

Combine the numerators over the common denominator.

$\frac{3 - 20}{289}$

Step 2.3

Subtract $20$ from $3$ .

$\frac{- 17}{289}$

Step 2.4

Cancel the common factor of $- 17$ and $289$ .

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Step 2.4.1

Factor $17$ out of $- 17$ .

$\frac{17 (- 1)}{289}$

Step 2.4.2

Cancel the common factors.

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Step 2.4.2.1

Factor $17$ out of $289$ .

$\frac{17 \cdot - 1}{17 \cdot 17}$

Step 2.4.2.2

Cancel the common factor.

$\frac{17 \cdot - 1}{17 \cdot 17}$

Step 2.4.2.3

Rewrite the expression.

$\frac{- 1}{17}$

Step 2.5

Move the negative in front of the fraction.

$- \frac{1}{17}$