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

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

Evaluate

$$25$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\left(-1\right)\left(-3\right)-1+3\times 2=11$$

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

$$-1+2\left(-1\right)\times 2-3\times 3=-14$$

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

$$11-\left(-14\right)$$

Subtract -14 from 11.

$$25$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\times 5-\left(-8\right)+7$$

Add the terms to obtain the final result.

$$25$$

Factor

$$5^{2}$$