Solve {lll}{0}&{1}&{0}{2}&{1}&{0}{3}&{3}&{2}

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

Evaluate

$$-4$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$\text{true}$$

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

$$2\times 2=4$$

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

$$-4$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-2\times 2$$

Simplify.

$$-4$$

Factor

$$-4$$