Solve {ccc}{2}&{3}&{1}{1}&{3}&{2}{3}&{2}&{1}

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

Evaluate

$$6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 3+3\times 2\times 3+2=26$$

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

$$3\times 3+2\times 2\times 2+3=20$$

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

$$26-20$$

Subtract 20 from 26.

$$6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-1\right)-3\left(-5\right)-7$$

Add the terms to obtain the final result.

$$6$$

Factor

$$2\times 3$$