Solve {lll}{2}&{1}&{1}{4}&{0}&{2}{2}&{2}&{0}

$$\left| \begin{array} { l l l } { 2 } & { 1 } & { 1 } \\ { 4 } & { 0 } & { 2 } \\ { 2 } & { 2 } & { 0 } \end{array} \right|$$

Evaluate

$$4$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&1&1\\4&0&2\\2&2&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&1&1&2&1\\4&0&2&4&0\\2&2&0&2&2\end{matrix}\right)$$

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

$$2\times 2+4\times 2=12$$

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

$$2\times 2\times 2=8$$

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

$$12-8$$

Subtract 8 from 12.

$$4$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&1&1\\4&0&2\\2&2&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}0&2\\2&0\end{matrix}\right))-det(\left(\begin{matrix}4&2\\2&0\end{matrix}\right))+det(\left(\begin{matrix}4&0\\2&2\end{matrix}\right))$$

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

$$2\left(-2\times 2\right)-\left(-2\times 2\right)+4\times 2$$

Simplify.

$$2\left(-4\right)-\left(-4\right)+8$$

Add the terms to obtain the final result.

$$4$$

Factor

$$2^{2}$$