Solve {ccc}{1}&{2}&{3}{4}&{-1}&{6}{5}&{6}&{2}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { 3 } \\ { 4 } & { - 1 } & { 6 } \\ { 5 } & { 6 } & { 2 } \end{array} \right|$$

Evaluate

$$93$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2+2\times 6\times 5+3\times 4\times 6=130$$

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

$$5\left(-1\right)\times 3+6\times 6+2\times 4\times 2=37$$

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

$$130-37$$

Subtract 37 from 130.

$$93$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-2-6\times 6-2\left(4\times 2-5\times 6\right)+3\left(4\times 6-5\left(-1\right)\right)$$

Simplify.

$$-38-2\left(-22\right)+3\times 29$$

Add the terms to obtain the final result.

$$93$$

Factor

$$3\times 31$$