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

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

Evaluate

$$90$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$3\left(-6\right)\times 4-3\times 3\left(-2\right)=-54$$

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

$$36-\left(-54\right)$$

Subtract -54 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).

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

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.

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

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

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

Simplify.

$$-\left(-2\left(-3\right)\right)+4\times 24$$

Add the terms to obtain the final result.

$$90$$

Factor

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