Solve {rrr}{2}&{2}&{4}{6}&{0}&{5}{5}&{-2}&{-4}

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

Evaluate

$$70$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-2\times 5\times 2-4\times 6\times 2=-68$$

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

$$2-\left(-68\right)$$

Subtract -68 from 2.

$$70$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\times 10-2\left(-49\right)+4\left(-12\right)$$

Add the terms to obtain the final result.

$$70$$

Factor

$$2\times 5\times 7$$