Finite Math Examples

$[\begin{array}{c} 2 & 4 \\ 8 & 6 \end{array}]$

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 $[\begin{array}{c} 2 & 4 \\ 8 & 6 \end{array}]$ is $- 20$ .

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

$determinant [\begin{array}{c} 2 & 4 \\ 8 & 6 \end{array}] = | \begin{array}{c} 2 & 4 \\ 8 & 6 \end{array} |$

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

$(2) (6) - 8 \cdot 4$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $6$ .

$12 - 8 \cdot 4$

Multiply $- 8$ by $4$ .

$12 - 32$

Subtract $32$ from $12$ .

$- 20$

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

$\frac{1}{- 20} [\begin{array}{c} 6 & - (4) \\ - (8) & 2 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 6 & - (4) \\ - (8) & 2 \end{array}]$ .

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Rearrange $- (4)$ .

$\frac{1}{- 20} [\begin{array}{c} 6 & - 4 \\ - (8) & 2 \end{array}]$

Rearrange $- (8)$ .

$\frac{1}{- 20} [\begin{array}{c} 6 & - 4 \\ - 8 & 2 \end{array}]$

Multiply $\frac{1}{- 20}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{- 20} \cdot 6 & \frac{1}{- 20} \cdot - 4 \\ \frac{1}{- 20} \cdot - 8 & \frac{1}{- 20} \cdot 2 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} \frac{1}{- 20} \cdot 6 & \frac{1}{- 20} \cdot - 4 \\ \frac{1}{- 20} \cdot - 8 & \frac{1}{- 20} \cdot 2 \end{array}]$ .

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Rearrange $\frac{1}{- 20} \cdot 6$ .

$[\begin{array}{c} - \frac{3}{10} & \frac{1}{- 20} \cdot - 4 \\ \frac{1}{- 20} \cdot - 8 & \frac{1}{- 20} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 20} \cdot - 4$ .

$[\begin{array}{c} - \frac{3}{10} & \frac{1}{5} \\ \frac{1}{- 20} \cdot - 8 & \frac{1}{- 20} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 20} \cdot - 8$ .

$[\begin{array}{c} - \frac{3}{10} & \frac{1}{5} \\ \frac{2}{5} & \frac{1}{- 20} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 20} \cdot 2$ .

$[\begin{array}{c} - \frac{3}{10} & \frac{1}{5} \\ \frac{2}{5} & - \frac{1}{10} \end{array}]$

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