Solve {ccc}{1}&{2}&{-1}{2}&{-3}&{2}{2}&{1}&{-5}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { - 1 } \\ { 2 } & { - 3 } & { 2 } \\ { 2 } & { 1 } & { - 5 } \end{array} \right|$$

Evaluate

$$33$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&-1\\2&-3&2\\2&1&-5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&-1&1&2\\2&-3&2&2&-3\\2&1&-5&2&1\end{matrix}\right)$$

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

$$-3\left(-5\right)+2\times 2\times 2-2=21$$

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

$$2\left(-3\right)\left(-1\right)+2-5\times 2\times 2=-12$$

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

$$21-\left(-12\right)$$

Subtract -12 from 21.

$$33$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&-1\\2&-3&2\\2&1&-5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-3&2\\1&-5\end{matrix}\right))-2det(\left(\begin{matrix}2&2\\2&-5\end{matrix}\right))-det(\left(\begin{matrix}2&-3\\2&1\end{matrix}\right))$$

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

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

Simplify.

$$13-2\left(-14\right)-8$$

Add the terms to obtain the final result.

$$33$$

Factor

$$3\times 11$$