Solve {ccc}{-1}&{0}&{1}{0}&{2}&{3}{-1}&{-1}&{-2}

$$\left| \begin{array} { c c c } { - 1 } & { 0 } & { 1 } \\ { 0 } & { 2 } & { 3 } \\ { - 1 } & { - 1 } & { - 2 } \end{array} \right|$$

Evaluate

$$3$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-1&0&1\\0&2&3\\-1&-1&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}-1&0&1&-1&0\\0&2&3&0&2\\-1&-1&-2&-1&-1\end{matrix}\right)$$

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

$$-2\left(-2\right)=4$$

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

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

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

$$4-1$$

Subtract 1 from 4.

$$3$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-1&0&1\\0&2&3\\-1&-1&-2\end{matrix}\right))$$

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.

$$-det(\left(\begin{matrix}2&3\\-1&-2\end{matrix}\right))+det(\left(\begin{matrix}0&2\\-1&-1\end{matrix}\right))$$

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

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

Simplify.

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

Add the terms to obtain the final result.

$$3$$

Factor

$$3$$