Solve {rrr}{1}&{-1}&{-2}{1}&{-3}&{-1}{1}&{2}&{2}

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

Evaluate

$$-11$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3\times 2-\left(-1\right)-2\times 2=-9$$

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

$$-3\left(-2\right)+2\left(-1\right)+2\left(-1\right)=2$$

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

$$-9-2$$

Subtract 2 from -9.

$$-11$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-4-\left(-3\right)-2\times 5$$

Add the terms to obtain the final result.

$$-11$$

Factor

$$-11$$