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

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

Evaluate

$$117$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 3\times 5+5\left(-2\right)\left(-4\right)=115$$

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

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

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

$$115-\left(-2\right)$$

Subtract -2 from 115.

$$117$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$5\times 25-5\left(-8\right)-4\times 12$$

Add the terms to obtain the final result.

$$117$$

Factor

$$3^{2}\times 13$$