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

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

Evaluate

$$12$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 4\times 6+5\times 8\times 3+2\times 6=180$$

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

$$3\times 4+6\times 8\times 2+6\times 2\times 5=168$$

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

$$180-168$$

Subtract 168 from 180.

$$12$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-24\right)-5\left(-12\right)$$

Add the terms to obtain the final result.

$$12$$

Factor

$$2^{2}\times 3$$