Solve {lll}{3}&{3}&{2}{1}&{2}&{1}{2}&{5}&{3}

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

Evaluate

$$2$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 2\times 3+3\times 2+2\times 5=34$$

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

$$2\times 2\times 2+5\times 3+3\times 3=32$$

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

$$34-32$$

Subtract 32 from 34.

$$2$$

Steps Using Expansion by Minors

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

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

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.

$$3det(\left(\begin{matrix}2&1\\5&3\end{matrix}\right))-3det(\left(\begin{matrix}1&1\\2&3\end{matrix}\right))+2det(\left(\begin{matrix}1&2\\2&5\end{matrix}\right))$$

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

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

Simplify.

$$3-3+2$$

Add the terms to obtain the final result.

$$2$$

Factor

$$2$$