Solve {ccc}{1}&{-4}&{-6}{-2}&{5}&{2}{-6}&{6}&{2}

$$\left| \begin{array} { c c c } { 1 } & { - 4 } & { - 6 } \\ { - 2 } & { 5 } & { 2 } \\ { - 6 } & { 6 } & { 2 } \end{array} \right|$$

Evaluate

$$-78$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-4&-6\\-2&5&2\\-6&6&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-4&-6&1&-4\\-2&5&2&-2&5\\-6&6&2&-6&6\end{matrix}\right)$$

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

$$5\times 2-4\times 2\left(-6\right)-6\left(-2\right)\times 6=130$$

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

$$-6\times 5\left(-6\right)+6\times 2+2\left(-2\right)\left(-4\right)=208$$

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

$$130-208$$

Subtract 208 from 130.

$$-78$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-4&-6\\-2&5&2\\-6&6&2\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}5&2\\6&2\end{matrix}\right))-\left(-4det(\left(\begin{matrix}-2&2\\-6&2\end{matrix}\right))\right)-6det(\left(\begin{matrix}-2&5\\-6&6\end{matrix}\right))$$

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

$$5\times 2-6\times 2-\left(-4\left(-2\times 2-\left(-6\times 2\right)\right)\right)-6\left(-2\times 6-\left(-6\times 5\right)\right)$$

Simplify.

$$-2-\left(-4\times 8\right)-6\times 18$$

Add the terms to obtain the final result.

$$-78$$

Factor

$$-78$$