Solve {ccc}{-3}&{0}&{-6}{2}&{-1}&{2}{3}&{-2}&{5}

$$\left| \begin{array} { c c c } { - 3 } & { 0 } & { - 6 } \\ { 2 } & { - 1 } & { 2 } \\ { 3 } & { - 2 } & { 5 } \end{array} \right|$$

Evaluate

$$9$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-3&0&-6\\2&-1&2\\3&-2&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}-3&0&-6&-3&0\\2&-1&2&2&-1\\3&-2&5&3&-2\end{matrix}\right)$$

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

$$-3\left(-1\right)\times 5-6\times 2\left(-2\right)=39$$

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

$$3\left(-1\right)\left(-6\right)-2\times 2\left(-3\right)=30$$

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

$$39-30$$

Subtract 30 from 39.

$$9$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-3&0&-6\\2&-1&2\\3&-2&5\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}-1&2\\-2&5\end{matrix}\right))-6det(\left(\begin{matrix}2&-1\\3&-2\end{matrix}\right))$$

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

$$-3\left(-5-\left(-2\times 2\right)\right)-6\left(2\left(-2\right)-3\left(-1\right)\right)$$

Simplify.

$$-3\left(-1\right)-6\left(-1\right)$$

Add the terms to obtain the final result.

$$9$$

Factor

$$3^{2}$$