Solve {ccc}{1}&{2}&{3}{3}&{1}&{0}{5}&{2}&{1}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { 3 } \\ { 3 } & { 1 } & { 0 } \\ { 5 } & { 2 } & { 1 } \end{array} \right|$$

Evaluate

$$-2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\3&1&0\\5&2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\3&1&0&3&1\\5&2&1&5&2\end{matrix}\right)$$

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

$$1+3\times 3\times 2=19$$

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

$$5\times 3+3\times 2=21$$

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

$$19-21$$

Subtract 21 from 19.

$$-2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\3&1&0\\5&2&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}1&0\\2&1\end{matrix}\right))-2det(\left(\begin{matrix}3&0\\5&1\end{matrix}\right))+3det(\left(\begin{matrix}3&1\\5&2\end{matrix}\right))$$

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

$$1-2\times 3+3\left(3\times 2-5\right)$$

Simplify.

$$1-2\times 3+3$$

Add the terms to obtain the final result.

$$-2$$

Factor

$$-2$$