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

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

Evaluate

$$-32$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 2+2\times 4+3\times 6\times 2=54$$

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

$$4\times 5\times 3+2+2\times 6\times 2=86$$

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

$$54-86$$

Subtract 86 from 54.

$$-32$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-32$$

Factor

$$-32$$