Solve {lll}{3}&{6}&{3}{1}&{3}&{-2}{2}&{2}&{2}

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

Evaluate

$$-18$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\times 2+6\left(-2\right)\times 2+3\times 2=0$$

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

$$2\times 3\times 3+2\left(-2\right)\times 3+2\times 6=18$$

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

$$-18$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 10-6\times 6+3\left(-4\right)$$

Add the terms to obtain the final result.

$$-18$$

Factor

$$-18$$