Solve {ccc}{3}&{2}&{4}{2}&{-2}&{2}{-7}&{6}&{3}

$$\left| \begin{array} { c c c } { 3 } & { 2 } & { 4 } \\ { 2 } & { - 2 } & { 2 } \\ { - 7 } & { 6 } & { 3 } \end{array} \right|$$

Evaluate

$$-102$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}3&2&4\\2&-2&2\\-7&6&3\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}3&2&4&3&2\\2&-2&2&2&-2\\-7&6&3&-7&6\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$3\left(-2\right)\times 3+2\times 2\left(-7\right)+4\times 2\times 6=2$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$-7\left(-2\right)\times 4+6\times 2\times 3+3\times 2\times 2=104$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$2-104$$

Subtract 104 from 2.

$$-102$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}3&2&4\\2&-2&2\\-7&6&3\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$3det(\left(\begin{matrix}-2&2\\6&3\end{matrix}\right))-2det(\left(\begin{matrix}2&2\\-7&3\end{matrix}\right))+4det(\left(\begin{matrix}2&-2\\-7&6\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$3\left(-2\times 3-6\times 2\right)-2\left(2\times 3-\left(-7\times 2\right)\right)+4\left(2\times 6-\left(-7\left(-2\right)\right)\right)$$

Simplify.

$$3\left(-18\right)-2\times 20+4\left(-2\right)$$

Add the terms to obtain the final result.

$$-102$$

Factor

$$-102$$