Solve {ccc}{3}&{4}&{2}{1}&{-4}&{-6}{-3}&{6}&{8}

$$\left| \begin{array} { c c c } { 3 } & { 4 } & { 2 } \\ { 1 } & { - 4 } & { - 6 } \\ { - 3 } & { 6 } & { 8 } \end{array} \right|$$

Evaluate

$$40$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&4&2\\1&-4&-6\\-3&6&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&4&2&3&4\\1&-4&-6&1&-4\\-3&6&8&-3&6\end{matrix}\right)$$

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

$$3\left(-4\right)\times 8+4\left(-6\right)\left(-3\right)+2\times 6=-12$$

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

$$-3\left(-4\right)\times 2+6\left(-6\right)\times 3+8\times 4=-52$$

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

$$-12-\left(-52\right)$$

Subtract -52 from -12.

$$40$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&4&2\\1&-4&-6\\-3&6&8\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}-4&-6\\6&8\end{matrix}\right))-4det(\left(\begin{matrix}1&-6\\-3&8\end{matrix}\right))+2det(\left(\begin{matrix}1&-4\\-3&6\end{matrix}\right))$$

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

$$3\left(-4\times 8-6\left(-6\right)\right)-4\left(8-\left(-3\left(-6\right)\right)\right)+2\left(6-\left(-3\left(-4\right)\right)\right)$$

Simplify.

$$3\times 4-4\left(-10\right)+2\left(-6\right)$$

Add the terms to obtain the final result.

$$40$$

Factor

$$2^{3}\times 5$$