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

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

Evaluate

$$-12$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2\times 2\times 3-1=11$$

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

$$-1-11$$

Subtract 11 from -1.

$$-12$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\left(-3\right)+3\left(-5\right)$$

Add the terms to obtain the final result.

$$-12$$

Factor

$$-12$$