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

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

Evaluate

$$-60$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-6\times 2-6\times 3-1=-31$$

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

$$3\times 6-1+2\left(-1\right)\left(-6\right)=29$$

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

$$-31-29$$

Subtract 29 from -31.

$$-60$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-11-\left(-6\left(-5\right)\right)-19$$

Add the terms to obtain the final result.

$$-60$$

Factor

$$-60$$