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

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

Evaluate

$$14$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-\left(-1\right)+3\times 3\times 3-2\times 2\times 2=20$$

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

$$3\left(-1\right)\left(-2\right)+2\times 3-2\times 3=6$$

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

$$20-6$$

Subtract 6 from 20.

$$14$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-\left(-1\right)-2\times 3-3\left(2\left(-1\right)-3\times 3\right)-2\left(2\times 2-3\left(-1\right)\right)$$

Simplify.

$$-5-3\left(-11\right)-2\times 7$$

Add the terms to obtain the final result.

$$14$$

Factor

$$2\times 7$$