Solve {rrr}{-2}&{0}&{-1}{1}&{-2}&{-2}{2}&{-2}&{-2}

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

Evaluate

$$-2$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\left(-2\right)\left(-2\right)-\left(-2\right)=-6$$

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

$$2\left(-2\right)\left(-1\right)-2\left(-2\right)\left(-2\right)=-4$$

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

$$-6-\left(-4\right)$$

Subtract -4 from -6.

$$-2$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-2\left(-2\left(-2\right)-\left(-2\left(-2\right)\right)\right)-\left(-2-2\left(-2\right)\right)$$

Simplify.

$$-2$$

Factor

$$-2$$