Solve {ccc}{3}&{-4}&{2}{6}&{1}&{3}{-4}&{-6}&{1}

$$\left| \begin{array} { c c c } { 3 } & { - 4 } & { 2 } \\ { 6 } & { 1 } & { 3 } \\ { - 4 } & { - 6 } & { 1 } \end{array} \right|$$

Evaluate

$$65$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&-4&2\\6&1&3\\-4&-6&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&-4&2&3&-4\\6&1&3&6&1\\-4&-6&1&-4&-6\end{matrix}\right)$$

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

$$3-4\times 3\left(-4\right)+2\times 6\left(-6\right)=-21$$

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

$$-4\times 2-6\times 3\times 3+6\left(-4\right)=-86$$

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

$$-21-\left(-86\right)$$

Subtract -86 from -21.

$$65$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&-4&2\\6&1&3\\-4&-6&1\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}1&3\\-6&1\end{matrix}\right))-\left(-4det(\left(\begin{matrix}6&3\\-4&1\end{matrix}\right))\right)+2det(\left(\begin{matrix}6&1\\-4&-6\end{matrix}\right))$$

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

$$3\left(1-\left(-6\times 3\right)\right)-\left(-4\left(6-\left(-4\times 3\right)\right)\right)+2\left(6\left(-6\right)-\left(-4\right)\right)$$

Simplify.

$$3\times 19-\left(-4\times 18\right)+2\left(-32\right)$$

Add the terms to obtain the final result.

$$65$$

Factor

$$5\times 13$$