Linear Algebra Examples

$\frac{1}{- 5} \cdot [\begin{array}{c} 19 & 4 \\ 48 & - 7 \end{array}]$

Step 1

Move the negative in front of the fraction.

$- \frac{1}{5} \cdot [\begin{array}{c} 19 & 4 \\ 48 & - 7 \end{array}]$

Step 2

Multiply $- \frac{1}{5}$ by each element of the matrix.

$[\begin{array}{c} - \frac{1}{5} \cdot 19 & - \frac{1}{5} \cdot 4 \\ - \frac{1}{5} \cdot 48 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3

Simplify each element in the matrix.

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Step 3.1

Multiply $- \frac{1}{5} \cdot 19$ .

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Step 3.1.1

Multiply $19$ by $- 1$ .

$[\begin{array}{c} - 19 (\frac{1}{5}) & - \frac{1}{5} \cdot 4 \\ - \frac{1}{5} \cdot 48 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.1.2

Combine $- 19$ and $\frac{1}{5}$ .

$[\begin{array}{c} \frac{- 19}{5} & - \frac{1}{5} \cdot 4 \\ - \frac{1}{5} \cdot 48 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.2

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{19}{5} & - \frac{1}{5} \cdot 4 \\ - \frac{1}{5} \cdot 48 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.3

Multiply $- \frac{1}{5} \cdot 4$ .

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Step 3.3.1

Multiply $4$ by $- 1$ .

$[\begin{array}{c} - \frac{19}{5} & - 4 (\frac{1}{5}) \\ - \frac{1}{5} \cdot 48 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.3.2

Combine $- 4$ and $\frac{1}{5}$ .

$[\begin{array}{c} - \frac{19}{5} & \frac{- 4}{5} \\ - \frac{1}{5} \cdot 48 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.4

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{19}{5} & - \frac{4}{5} \\ - \frac{1}{5} \cdot 48 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.5

Multiply $- \frac{1}{5} \cdot 48$ .

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Step 3.5.1

Multiply $48$ by $- 1$ .

$[\begin{array}{c} - \frac{19}{5} & - \frac{4}{5} \\ - 48 (\frac{1}{5}) & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.5.2

Combine $- 48$ and $\frac{1}{5}$ .

$[\begin{array}{c} - \frac{19}{5} & - \frac{4}{5} \\ \frac{- 48}{5} & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.6

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{19}{5} & - \frac{4}{5} \\ - \frac{48}{5} & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 3.7

Multiply $- \frac{1}{5} \cdot - 7$ .

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Step 3.7.1

Multiply $- 7$ by $- 1$ .

$[\begin{array}{c} - \frac{19}{5} & - \frac{4}{5} \\ - \frac{48}{5} & 7 (\frac{1}{5}) \end{array}]$

Step 3.7.2

Combine $7$ and $\frac{1}{5}$ .

$[\begin{array}{c} - \frac{19}{5} & - \frac{4}{5} \\ - \frac{48}{5} & \frac{7}{5} \end{array}]$

Step 4

The inverse of a $2 \times 2$ matrix can be found using the formula $\frac{1}{a d - b c} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where $a d - b c$ is the determinant.

Step 5

Find the determinant.

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Step 5.1

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

$- \frac{19}{5} \cdot \frac{7}{5} - (- \frac{48}{5} (- \frac{4}{5}))$

Step 5.2

Simplify the determinant.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply $- \frac{19}{5} \cdot \frac{7}{5}$ .

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Step 5.2.1.1.1

Multiply $\frac{7}{5}$ by $\frac{19}{5}$ .

$- \frac{7 \cdot 19}{5 \cdot 5} - (- \frac{48}{5} (- \frac{4}{5}))$

Step 5.2.1.1.2

Multiply $7$ by $19$ .

$- \frac{133}{5 \cdot 5} - (- \frac{48}{5} (- \frac{4}{5}))$

Step 5.2.1.1.3

Multiply $5$ by $5$ .

$- \frac{133}{25} - (- \frac{48}{5} (- \frac{4}{5}))$

Step 5.2.1.2

Multiply $- \frac{48}{5} (- \frac{4}{5})$ .

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Step 5.2.1.2.1

Multiply $- 1$ by $- 1$ .

$- \frac{133}{25} - (1 (\frac{48}{5}) \frac{4}{5})$

Step 5.2.1.2.2

Multiply $\frac{48}{5}$ by $1$ .

$- \frac{133}{25} - (\frac{48}{5} \cdot \frac{4}{5})$

Step 5.2.1.2.3

Multiply $\frac{48}{5}$ by $\frac{4}{5}$ .

$- \frac{133}{25} - \frac{48 \cdot 4}{5 \cdot 5}$

Step 5.2.1.2.4

Multiply $48$ by $4$ .

$- \frac{133}{25} - \frac{192}{5 \cdot 5}$

Step 5.2.1.2.5

Multiply $5$ by $5$ .

$- \frac{133}{25} - \frac{192}{25}$

Step 5.2.2

Combine the numerators over the common denominator.

$\frac{- 133 - 192}{25}$

Step 5.2.3

Subtract $192$ from $- 133$ .

$\frac{- 325}{25}$

Step 5.2.4

Divide $- 325$ by $25$ .

$- 13$

Step 6

Since the determinant is non-zero, the inverse exists.

Step 7

Substitute the known values into the formula for the inverse.

$\frac{1}{- 13} [\begin{array}{c} \frac{7}{5} & \frac{4}{5} \\ \frac{48}{5} & - \frac{19}{5} \end{array}]$

Step 8

Move the negative in front of the fraction.

$- \frac{1}{13} [\begin{array}{c} \frac{7}{5} & \frac{4}{5} \\ \frac{48}{5} & - \frac{19}{5} \end{array}]$

Step 9

Multiply $- \frac{1}{13}$ by each element of the matrix.

$[\begin{array}{c} - \frac{1}{13} \cdot \frac{7}{5} & - \frac{1}{13} \cdot \frac{4}{5} \\ - \frac{1}{13} \cdot \frac{48}{5} & - \frac{1}{13} (- \frac{19}{5}) \end{array}]$

Step 10

Simplify each element in the matrix.

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Step 10.1

Multiply $- \frac{1}{13} \cdot \frac{7}{5}$ .

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Step 10.1.1

Multiply $\frac{7}{5}$ by $\frac{1}{13}$ .

$[\begin{array}{c} - \frac{7}{5 \cdot 13} & - \frac{1}{13} \cdot \frac{4}{5} \\ - \frac{1}{13} \cdot \frac{48}{5} & - \frac{1}{13} (- \frac{19}{5}) \end{array}]$

Step 10.1.2

Multiply $5$ by $13$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{1}{13} \cdot \frac{4}{5} \\ - \frac{1}{13} \cdot \frac{48}{5} & - \frac{1}{13} (- \frac{19}{5}) \end{array}]$

Step 10.2

Multiply $- \frac{1}{13} \cdot \frac{4}{5}$ .

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Step 10.2.1

Multiply $\frac{4}{5}$ by $\frac{1}{13}$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{5 \cdot 13} \\ - \frac{1}{13} \cdot \frac{48}{5} & - \frac{1}{13} (- \frac{19}{5}) \end{array}]$

Step 10.2.2

Multiply $5$ by $13$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{65} \\ - \frac{1}{13} \cdot \frac{48}{5} & - \frac{1}{13} (- \frac{19}{5}) \end{array}]$

Step 10.3

Multiply $- \frac{1}{13} \cdot \frac{48}{5}$ .

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Step 10.3.1

Multiply $\frac{48}{5}$ by $\frac{1}{13}$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{65} \\ - \frac{48}{5 \cdot 13} & - \frac{1}{13} (- \frac{19}{5}) \end{array}]$

Step 10.3.2

Multiply $5$ by $13$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{65} \\ - \frac{48}{65} & - \frac{1}{13} (- \frac{19}{5}) \end{array}]$

Step 10.4

Multiply $- \frac{1}{13} (- \frac{19}{5})$ .

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Step 10.4.1

Multiply $- 1$ by $- 1$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{65} \\ - \frac{48}{65} & 1 (\frac{1}{13}) \frac{19}{5} \end{array}]$

Step 10.4.2

Multiply $\frac{1}{13}$ by $1$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{65} \\ - \frac{48}{65} & \frac{1}{13} \cdot \frac{19}{5} \end{array}]$

Step 10.4.3

Multiply $\frac{1}{13}$ by $\frac{19}{5}$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{65} \\ - \frac{48}{65} & \frac{19}{13 \cdot 5} \end{array}]$

Step 10.4.4

Multiply $13$ by $5$ .

$[\begin{array}{c} - \frac{7}{65} & - \frac{4}{65} \\ - \frac{48}{65} & \frac{19}{65} \end{array}]$