Find the Inverse of the Matrix [[40/13,70/13],[28/13,140/13]]

$[\begin{array}{c} \frac{40}{13} & \frac{70}{13} \\ \frac{28}{13} & \frac{140}{13} \end{array}]$

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

If $A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then $A^{- 1} = \frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

The determinant of $[\begin{array}{c} \frac{40}{13} & \frac{70}{13} \\ \frac{28}{13} & \frac{140}{13} \end{array}]$ is $\frac{280}{13}$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} \frac{40}{13} & \frac{70}{13} \\ \frac{28}{13} & \frac{140}{13} \end{array}] = | \begin{array}{c} \frac{40}{13} & \frac{70}{13} \\ \frac{28}{13} & \frac{140}{13} \end{array} |$

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

$(\frac{40}{13}) (\frac{140}{13}) - \frac{28}{13} \cdot \frac{70}{13}$

Simplify the determinant.

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Simplify each term.

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Multiply $(\frac{40}{13}) (\frac{140}{13})$ .

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Multiply $\frac{40}{13}$ and $\frac{140}{13}$ .

$\frac{40 \cdot 140}{13 \cdot 13} - \frac{28}{13} \cdot \frac{70}{13}$

Multiply $40$ by $140$ .

$\frac{5600}{13 \cdot 13} - \frac{28}{13} \cdot \frac{70}{13}$

Multiply $13$ by $13$ .

$\frac{5600}{169} - \frac{28}{13} \cdot \frac{70}{13}$

Multiply $- \frac{28}{13} \cdot \frac{70}{13}$ .

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Multiply $\frac{70}{13}$ and $\frac{28}{13}$ .

$\frac{5600}{169} - \frac{70 \cdot 28}{13 \cdot 13}$

Multiply $70$ by $28$ .

$\frac{5600}{169} - \frac{1960}{13 \cdot 13}$

Multiply $13$ by $13$ .

$\frac{5600}{169} - \frac{1960}{169}$

Combine the numerators over the common denominator.

$\frac{5600 - 1960}{169}$

Subtract $1960$ from $5600$ .

$\frac{3640}{169}$

Cancel the common factor of $3640$ and $169$ .

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Factor $13$ out of $3640$ .

$\frac{13 (280)}{169}$

Cancel the common factors.

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Factor $13$ out of $169$ .

$\frac{13 \cdot 280}{13 \cdot 13}$

Cancel the common factor.

$\frac{13 \cdot 280}{13 \cdot 13}$

Rewrite the expression.

$\frac{280}{13}$

Substitute the known values into the formula for the inverse of a matrix.

$\frac{1}{\frac{280}{13}} [\begin{array}{c} \frac{140}{13} & - (\frac{70}{13}) \\ - (\frac{28}{13}) & \frac{40}{13} \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} \frac{140}{13} & - (\frac{70}{13}) \\ - (\frac{28}{13}) & \frac{40}{13} \end{array}]$ .

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Rearrange $- (\frac{70}{13})$ .

$\frac{1}{\frac{280}{13}} [\begin{array}{c} \frac{140}{13} & - \frac{70}{13} \\ - (\frac{28}{13}) & \frac{40}{13} \end{array}]$

Rearrange $- (\frac{28}{13})$ .

$\frac{1}{\frac{280}{13}} [\begin{array}{c} \frac{140}{13} & - \frac{70}{13} \\ - \frac{28}{13} & \frac{40}{13} \end{array}]$

Multiply $\frac{1}{\frac{280}{13}}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{\frac{280}{13}} \cdot \frac{140}{13} & \frac{1}{\frac{280}{13}} \cdot (- \frac{70}{13}) \\ \frac{1}{\frac{280}{13}} \cdot (- \frac{28}{13}) & \frac{1}{\frac{280}{13}} \cdot \frac{40}{13} \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} \frac{1}{\frac{280}{13}} \cdot \frac{140}{13} & \frac{1}{\frac{280}{13}} \cdot (- \frac{70}{13}) \\ \frac{1}{\frac{280}{13}} \cdot (- \frac{28}{13}) & \frac{1}{\frac{280}{13}} \cdot \frac{40}{13} \end{array}]$ .

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Rearrange $\frac{1}{\frac{280}{13}} \cdot \frac{140}{13}$ .

$[\begin{array}{c} \frac{1}{2} & \frac{1}{\frac{280}{13}} \cdot (- \frac{70}{13}) \\ \frac{1}{\frac{280}{13}} \cdot (- \frac{28}{13}) & \frac{1}{\frac{280}{13}} \cdot \frac{40}{13} \end{array}]$

Rearrange $\frac{1}{\frac{280}{13}} \cdot (- \frac{70}{13})$ .

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{4} \\ \frac{1}{\frac{280}{13}} \cdot (- \frac{28}{13}) & \frac{1}{\frac{280}{13}} \cdot \frac{40}{13} \end{array}]$

Rearrange $\frac{1}{\frac{280}{13}} \cdot (- \frac{28}{13})$ .

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{4} \\ - \frac{1}{10} & \frac{1}{\frac{280}{13}} \cdot \frac{40}{13} \end{array}]$

Rearrange $\frac{1}{\frac{280}{13}} \cdot \frac{40}{13}$ .

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{4} \\ - \frac{1}{10} & \frac{1}{7} \end{array}]$

Find the Inverse of the Matrix [[40/13,70/13],[28/13,140/13]]

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