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

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

Evaluate

$$-5$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$3\left(-1\right)\times 2-3-2\left(-1\right)=-7$$

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

$$-12-\left(-7\right)$$

Subtract -7 from -12.

$$-5$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$4-\left(-\left(-11\right)\right)+2$$

Add the terms to obtain the final result.

$$-5$$

Factor

$$-5$$