Solve {ccc}{0}&{5}&{4}{5}&{6}&{-6}{-2}&{-3}&{2}

$$\left| \begin{array} { c c c } { 0 } & { 5 } & { 4 } \\ { 5 } & { 6 } & { - 6 } \\ { - 2 } & { - 3 } & { 2 } \end{array} \right|$$

Evaluate

$$-2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&5&4\\5&6&-6\\-2&-3&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&5&4&0&5\\5&6&-6&5&6\\-2&-3&2&-2&-3\end{matrix}\right)$$

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

$$5\left(-6\right)\left(-2\right)+4\times 5\left(-3\right)=0$$

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

$$-2\times 6\times 4+2\times 5\times 5=2$$

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

$$-2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&5&4\\5&6&-6\\-2&-3&2\end{matrix}\right))$$

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.

$$-5det(\left(\begin{matrix}5&-6\\-2&2\end{matrix}\right))+4det(\left(\begin{matrix}5&6\\-2&-3\end{matrix}\right))$$

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

$$-5\left(5\times 2-\left(-2\left(-6\right)\right)\right)+4\left(5\left(-3\right)-\left(-2\times 6\right)\right)$$

Simplify.

$$-5\left(-2\right)+4\left(-3\right)$$

Add the terms to obtain the final result.

$$-2$$

Factor

$$-2$$