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

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

Evaluate

$$-64$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\left(-2\right)\left(-3\right)-3\times 4\times 2=-12$$

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

$$-3\times 4\left(-3\right)+2\times 4\times 2=52$$

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

$$-12-52$$

Subtract 52 from -12.

$$-64$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2\times 2-3\times 20$$

Add the terms to obtain the final result.

$$-64$$

Factor

$$-64$$