Solve {ccc}{-3}&{6}&{9}{15}&{3}&{0}{6}&{9}&{-9}

$$\left| \begin{array} { c c c } { - 3 } & { 6 } & { 9 } \\ { 15 } & { 3 } & { 0 } \\ { 6 } & { 9 } & { - 9 } \end{array} \right|$$

Evaluate

$$1944$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-3&6&9\\15&3&0\\6&9&-9\end{matrix}\right))$$

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

$$\left(\begin{matrix}-3&6&9&-3&6\\15&3&0&15&3\\6&9&-9&6&9\end{matrix}\right)$$

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

$$-3\times 3\left(-9\right)+9\times 15\times 9=1296$$

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

$$6\times 3\times 9-9\times 15\times 6=-648$$

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

$$1296-\left(-648\right)$$

Subtract -648 from 1296.

$$1944$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-3&6&9\\15&3&0\\6&9&-9\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}3&0\\9&-9\end{matrix}\right))-6det(\left(\begin{matrix}15&0\\6&-9\end{matrix}\right))+9det(\left(\begin{matrix}15&3\\6&9\end{matrix}\right))$$

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

$$-3\times 3\left(-9\right)-6\times 15\left(-9\right)+9\left(15\times 9-6\times 3\right)$$

Simplify.

$$-3\left(-27\right)-6\left(-135\right)+9\times 117$$

Add the terms to obtain the final result.

$$1944$$

Factor

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