Solve {rrr}{3}&{3}&{-2}{-1}&{3}&{2}{2}&{3}&{-2}

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

Evaluate

$$-12$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\left(-2\right)+3\times 2\times 2-2\left(-1\right)\times 3=0$$

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

$$2\times 3\left(-2\right)+3\times 2\times 3-2\left(-1\right)\times 3=12$$

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

$$-12$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-12\right)-3\left(-2\right)-2\left(-9\right)$$

Add the terms to obtain the final result.

$$-12$$

Factor

$$-12$$