Solve {ccc}{3}&{3}&{-6}{-3}&{0}&{6}{2}&{-4}&{6}

$$\left| \begin{array} { c c c } { 3 } & { 3 } & { - 6 } \\ { - 3 } & { 0 } & { 6 } \\ { 2 } & { - 4 } & { 6 } \end{array} \right|$$

Evaluate

$$90$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&3&-6\\-3&0&6\\2&-4&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&3&-6&3&3\\-3&0&6&-3&0\\2&-4&6&2&-4\end{matrix}\right)$$

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

$$3\times 6\times 2-6\left(-3\right)\left(-4\right)=-36$$

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

$$-4\times 6\times 3+6\left(-3\right)\times 3=-126$$

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

$$-36-\left(-126\right)$$

Subtract -126 from -36.

$$90$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&3&-6\\-3&0&6\\2&-4&6\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}0&6\\-4&6\end{matrix}\right))-3det(\left(\begin{matrix}-3&6\\2&6\end{matrix}\right))-6det(\left(\begin{matrix}-3&0\\2&-4\end{matrix}\right))$$

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

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

Simplify.

$$3\times 24-3\left(-30\right)-6\times 12$$

Add the terms to obtain the final result.

$$90$$

Factor

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