Finite Math Examples

$[\begin{array}{c} 6 & 8 \\ 2 & 5 \end{array}]$

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 $[\begin{array}{c} 6 & 8 \\ 2 & 5 \end{array}]$ is $14$ .

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

$determinant [\begin{array}{c} 6 & 8 \\ 2 & 5 \end{array}] = | \begin{array}{c} 6 & 8 \\ 2 & 5 \end{array} |$

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

$(6) (5) - 2 \cdot 8$

Simplify the determinant.

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Simplify each term.

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Multiply $6$ by $5$ .

$30 - 2 \cdot 8$

Multiply $- 2$ by $8$ .

$30 - 16$

Subtract $16$ from $30$ .

$14$

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

$\frac{1}{14} [\begin{array}{c} 5 & - (8) \\ - (2) & 6 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 5 & - (8) \\ - (2) & 6 \end{array}]$ .

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Simplify element $0, 1$ by multiplying $- (8)$ .

$\frac{1}{14} [\begin{array}{c} 5 & - 8 \\ - (2) & 6 \end{array}]$

Simplify element $1, 0$ by multiplying $- (2)$ .

$\frac{1}{14} [\begin{array}{c} 5 & - 8 \\ - 2 & 6 \end{array}]$

Multiply $\frac{1}{14}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{14} \cdot 5 & \frac{1}{14} \cdot - 8 \\ \frac{1}{14} \cdot - 2 & \frac{1}{14} \cdot 6 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} \frac{1}{14} \cdot 5 & \frac{1}{14} \cdot - 8 \\ \frac{1}{14} \cdot - 2 & \frac{1}{14} \cdot 6 \end{array}]$ .

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Simplify element $0, 0$ by multiplying $\frac{1}{14} \cdot 5$ .

$[\begin{array}{c} \frac{5}{14} & \frac{1}{14} \cdot - 8 \\ \frac{1}{14} \cdot - 2 & \frac{1}{14} \cdot 6 \end{array}]$

Simplify element $0, 1$ by multiplying $\frac{1}{14} \cdot - 8$ .

$[\begin{array}{c} \frac{5}{14} & - \frac{4}{7} \\ \frac{1}{14} \cdot - 2 & \frac{1}{14} \cdot 6 \end{array}]$

Simplify element $1, 0$ by multiplying $\frac{1}{14} \cdot - 2$ .

$[\begin{array}{c} \frac{5}{14} & - \frac{4}{7} \\ - \frac{1}{7} & \frac{1}{14} \cdot 6 \end{array}]$

Simplify element $1, 1$ by multiplying $\frac{1}{14} \cdot 6$ .

$[\begin{array}{c} \frac{5}{14} & - \frac{4}{7} \\ - \frac{1}{7} & \frac{3}{7} \end{array}]$

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