Algebra Examples

$[\begin{array}{c} - 1 & 0 \\ 0 & - 5 \end{array}]$

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

Find the determinant of $[\begin{array}{c} - 1 & 0 \\ 0 & - 5 \end{array}]$ .

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

$determinant [\begin{array}{c} - 1 & 0 \\ 0 & - 5 \end{array}] = | \begin{array}{c} - 1 & 0 \\ 0 & - 5 \end{array} |$

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

$(- 1) (- 5) + 0 \cdot 0$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $- 5$ .

$5 + 0 \cdot 0$

Multiply $0$ by $0$ .

$5 + 0$

Add $5$ and $0$ .

$5$

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

$\frac{1}{5} [\begin{array}{c} - 5 & - (0) \\ - (0) & - 1 \end{array}]$

Simplify each element in the matrix.

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Rearrange $- (0)$ .

$\frac{1}{5} [\begin{array}{c} - 5 & 0 \\ - (0) & - 1 \end{array}]$

Rearrange $- (0)$ .

$\frac{1}{5} [\begin{array}{c} - 5 & 0 \\ 0 & - 1 \end{array}]$

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

$[\begin{array}{c} \frac{1}{5} \cdot - 5 & \frac{1}{5} \cdot 0 \\ \frac{1}{5} \cdot 0 & \frac{1}{5} \cdot - 1 \end{array}]$

Simplify each element in the matrix.

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

$[\begin{array}{c} - 1 & \frac{1}{5} \cdot 0 \\ \frac{1}{5} \cdot 0 & \frac{1}{5} \cdot - 1 \end{array}]$

Rearrange $\frac{1}{5} \cdot 0$ .

$[\begin{array}{c} - 1 & 0 \\ \frac{1}{5} \cdot 0 & \frac{1}{5} \cdot - 1 \end{array}]$

Rearrange $\frac{1}{5} \cdot 0$ .

$[\begin{array}{c} - 1 & 0 \\ 0 & \frac{1}{5} \cdot - 1 \end{array}]$

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

$[\begin{array}{c} - 1 & 0 \\ 0 & - \frac{1}{5} \end{array}]$