Finite Math Examples

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

Step 1

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

Step 2

Find the determinant of $[\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}]$ .

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

$determinant [\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] = | \begin{array}{c} 1 & 2 \\ 3 & 5 \end{array} |$

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

$(1) (5) - 3 \cdot 2$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $1$ .

$5 - 3 \cdot 2$

Multiply $- 3$ by $2$ .

$5 - 6$

Subtract $6$ from $5$ .

$- 1$

Step 3

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

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

Step 4

Simplify each element in the matrix.

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Rearrange $- (2)$ .

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

Rearrange $- (3)$ .

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

Step 5

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

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

Step 6

Simplify each element in the matrix.

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Rearrange $\frac{1}{- 1} \cdot 5$ .

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

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

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

Rearrange $\frac{1}{- 1} \cdot - 3$ .

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

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

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

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