Precalculus Examples

$[\begin{array}{c} 2 & 4 \\ - 2 & - 4 \end{array}] + [\begin{array}{c} 4 & - 1 \\ 3 & 2 \end{array}]$

Add the corresponding elements.

$[\begin{array}{c} 4 + 2 & - 1 + 4 \\ 3 - 2 & 2 - 4 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 4 + 2 & - 1 + 4 \\ 3 - 2 & 2 - 4 \end{array}]$ .

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Simplify $4 + 2$ .

$[\begin{array}{c} 6 & - 1 + 4 \\ 3 - 2 & 2 - 4 \end{array}]$

Simplify $- 1 + 4$ .

$[\begin{array}{c} 6 & 3 \\ 3 - 2 & 2 - 4 \end{array}]$

Simplify $3 - 2$ .

$[\begin{array}{c} 6 & 3 \\ 1 & 2 - 4 \end{array}]$

Simplify $2 - 4$ .

$[\begin{array}{c} 6 & 3 \\ 1 & - 2 \end{array}]$

Find the determinant of $[\begin{array}{c} 6 & 3 \\ 1 & - 2 \end{array}]$ .

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

$determinant [\begin{array}{c} 6 & 3 \\ 1 & - 2 \end{array}] = | \begin{array}{c} 6 & 3 \\ 1 & - 2 \end{array} |$

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

$(6) (- 2) - 1 \cdot 3$

Simplify the determinant.

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Simplify each term.

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Multiply $6$ by $- 2$ .

$- 12 - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$- 12 - 3$

Subtract $3$ from $- 12$ .

$- 15$

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