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

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

Evaluate

$$8$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2\times 2+4\times 2\times 5+3\times 6=66$$

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

$$5\times 2+6\times 2\times 2+2\times 3\times 4=58$$

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

$$66-58$$

Subtract 58 from 66.

$$8$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-8\right)-4\left(-4\right)+8$$

Add the terms to obtain the final result.

$$8$$

Factor

$$2^{3}$$