Solve {ccc}{1}&{0}&{-6}{-2}&{-8}&{9}{-2}&{-10}&{16}=

$$\left| \begin{array} { c c c } { 1 } & { 0 } & { - 6 } \\ { - 2 } & { - 8 } & { 9 } \\ { - 2 } & { - 10 } & { 16 } \end{array} \right| =$$

Evaluate

$$-62$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&-6\\-2&-8&9\\-2&-10&16\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&-6&1&0\\-2&-8&9&-2&-8\\-2&-10&16&-2&-10\end{matrix}\right)$$

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

$$-8\times 16-6\left(-2\right)\left(-10\right)=-248$$

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

$$-2\left(-8\right)\left(-6\right)-10\times 9=-186$$

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

$$-248-\left(-186\right)$$

Subtract -186 from -248.

$$-62$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&-6\\-2&-8&9\\-2&-10&16\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-8&9\\-10&16\end{matrix}\right))-6det(\left(\begin{matrix}-2&-8\\-2&-10\end{matrix}\right))$$

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

$$-8\times 16-\left(-10\times 9\right)-6\left(-2\left(-10\right)-\left(-2\left(-8\right)\right)\right)$$

Simplify.

$$-38-6\times 4$$

Add the terms to obtain the final result.

$$-62$$

Factor

$$-62$$