Solve {lll}{2}&{3}&{5}{6}&{7}&{8}{8}&{1}&{0}

$$\left| \begin{array} { l l l } { 2 } & { 3 } & { 5 } \\ { 6 } & { 7 } & { 8 } \\ { 8 } & { 1 } & { 0 } \end{array} \right|$$

Evaluate

$$-74$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&5\\6&7&8\\8&1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&5&2&3\\6&7&8&6&7\\8&1&0&8&1\end{matrix}\right)$$

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

$$3\times 8\times 8+5\times 6=222$$

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

$$8\times 7\times 5+8\times 2=296$$

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

$$222-296$$

Subtract 296 from 222.

$$-74$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&5\\6&7&8\\8&1&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}7&8\\1&0\end{matrix}\right))-3det(\left(\begin{matrix}6&8\\8&0\end{matrix}\right))+5det(\left(\begin{matrix}6&7\\8&1\end{matrix}\right))$$

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

$$2\left(-8\right)-3\left(-8\times 8\right)+5\left(6-8\times 7\right)$$

Simplify.

$$2\left(-8\right)-3\left(-64\right)+5\left(-50\right)$$

Add the terms to obtain the final result.

$$-74$$

Factor

$$-74$$