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

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

Evaluate

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2-3=-1$$

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

$$-7-\left(-1\right)$$

Subtract -1 from -7.

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-3-3$$

Add the terms to obtain the final result.

$$-6$$

Factor

$$-6$$