Solve {rrr}{2}&{3}&{1}{1}&{1}&{-6}{1}&{-2}&{6}

$$\left| \begin{array} { r r r } { 2 } & { 3 } & { 1 } \\ { 1 } & { 1 } & { - 6 } \\ { 1 } & { - 2 } & { 6 } \end{array} \right|$$

Evaluate

$$-51$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&1\\1&1&-6\\1&-2&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&1&2&3\\1&1&-6&1&1\\1&-2&6&1&-2\end{matrix}\right)$$

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

$$2\times 6+3\left(-6\right)-2=-8$$

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

$$1-2\left(-6\right)\times 2+6\times 3=43$$

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

$$-8-43$$

Subtract 43 from -8.

$$-51$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&1\\1&1&-6\\1&-2&6\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&-6\\-2&6\end{matrix}\right))-3det(\left(\begin{matrix}1&-6\\1&6\end{matrix}\right))+det(\left(\begin{matrix}1&1\\1&-2\end{matrix}\right))$$

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

$$2\left(6-\left(-2\left(-6\right)\right)\right)-3\left(6-\left(-6\right)\right)-2-1$$

Simplify.

$$2\left(-6\right)-3\times 12-3$$

Add the terms to obtain the final result.

$$-51$$

Factor

$$-51$$