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

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

Evaluate

$$-15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\left(-2\right)\left(-2\right)-2\left(-1\right)-1=-7$$

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

$$-2\left(-1\right)-\left(-2\right)-2\left(-2\right)=8$$

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

$$-7-8$$

Subtract 8 from -7.

$$-15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-15$$

Factor

$$-15$$