Solve {rrr}{1}&{1}&{-1}{1}&{2}&{0}{-1}&{0}&{-1}

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

Evaluate

$$-3$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

$$-2-1$$

Subtract 1 from -2.

$$-3$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-3$$

Factor

$$-3$$