Solve {ccc}{0}&{3}&{4}{3}&{0}&{6}{4}&{6}&{-8}

$$\left| \begin{array} { c c c } { 0 } & { 3 } & { 4 } \\ { 3 } & { 0 } & { 6 } \\ { 4 } & { 6 } & { - 8 } \end{array} \right|$$

Evaluate

$$216$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&3&4\\3&0&6\\4&6&-8\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&3&4&0&3\\3&0&6&3&0\\4&6&-8&4&6\end{matrix}\right)$$

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

$$3\times 6\times 4+4\times 3\times 6=144$$

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

$$-8\times 3\times 3=-72$$

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

$$144-\left(-72\right)$$

Subtract -72 from 144.

$$216$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&3&4\\3&0&6\\4&6&-8\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}3&6\\4&-8\end{matrix}\right))+4det(\left(\begin{matrix}3&0\\4&6\end{matrix}\right))$$

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

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

Simplify.

$$-3\left(-48\right)+4\times 18$$

Add the terms to obtain the final result.

$$216$$

Factor

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