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

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

Evaluate

$$8$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

$$3-\left(-5\right)$$

Subtract -5 from 3.

$$8$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-1-\left(-\left(-3\right)\right)+3\times 4$$

Add the terms to obtain the final result.

$$8$$

Factor

$$2^{3}$$