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

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

Evaluate

$$3$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 3=6$$

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

$$3=3$$

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

$$6-3$$

Subtract 3 from 6.

$$3$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-3+2\times 3$$

Add the terms to obtain the final result.

$$3$$

Factor

$$3$$