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

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

Evaluate

$$-10$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3\times 2-2\times 2-1=-11$$

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

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

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

$$-11-\left(-1\right)$$

Subtract -1 from -11.

$$-10$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-5-5$$

Add the terms to obtain the final result.

$$-10$$

Factor

$$-10$$