Solve {ccc}{1}&{2}&{-4}{-2}&{2}&{1}{-3}&{4}&{-2}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { - 4 } \\ { - 2 } & { 2 } & { 1 } \\ { - 3 } & { 4 } & { - 2 } \end{array} \right|$$

Evaluate

$$-14$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&-4\\-2&2&1\\-3&4&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&-4&1&2\\-2&2&1&-2&2\\-3&4&-2&-3&4\end{matrix}\right)$$

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

$$2\left(-2\right)+2\left(-3\right)-4\left(-2\right)\times 4=22$$

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

$$-3\times 2\left(-4\right)+4-2\left(-2\right)\times 2=36$$

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

$$22-36$$

Subtract 36 from 22.

$$-14$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&-4\\-2&2&1\\-3&4&-2\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}2&1\\4&-2\end{matrix}\right))-2det(\left(\begin{matrix}-2&1\\-3&-2\end{matrix}\right))-4det(\left(\begin{matrix}-2&2\\-3&4\end{matrix}\right))$$

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

$$2\left(-2\right)-4-2\left(-2\left(-2\right)-\left(-3\right)\right)-4\left(-2\times 4-\left(-3\times 2\right)\right)$$

Simplify.

$$-8-2\times 7-4\left(-2\right)$$

Add the terms to obtain the final result.

$$-14$$

Factor

$$-14$$