Solve {lll}{5}&{0}&{2}{4}&{1}&{5}{3}&{8}&{6}

$$\left| \begin{array} { l l l } { 5 } & { 0 } & { 2 } \\ { 4 } & { 1 } & { 5 } \\ { 3 } & { 8 } & { 6 } \end{array} \right|$$

Evaluate

$$-112$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&0&2\\4&1&5\\3&8&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&0&2&5&0\\4&1&5&4&1\\3&8&6&3&8\end{matrix}\right)$$

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

$$5\times 6+2\times 4\times 8=94$$

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

$$3\times 2+8\times 5\times 5=206$$

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

$$94-206$$

Subtract 206 from 94.

$$-112$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&0&2\\4&1&5\\3&8&6\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}1&5\\8&6\end{matrix}\right))+2det(\left(\begin{matrix}4&1\\3&8\end{matrix}\right))$$

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

$$5\left(6-8\times 5\right)+2\left(4\times 8-3\right)$$

Simplify.

$$5\left(-34\right)+2\times 29$$

Add the terms to obtain the final result.

$$-112$$

Factor

$$-112$$