Solve {ccc}{-2}&{3}&{4}{-5}&{1}&{-5}{-5}&{-2}&{1}

$$\left| \begin{array} { c c c } { - 2 } & { 3 } & { 4 } \\ { - 5 } & { 1 } & { - 5 } \\ { - 5 } & { - 2 } & { 1 } \end{array} \right|$$

Evaluate

$$168$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&3&4\\-5&1&-5\\-5&-2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&3&4&-2&3\\-5&1&-5&-5&1\\-5&-2&1&-5&-2\end{matrix}\right)$$

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

$$-2+3\left(-5\right)\left(-5\right)+4\left(-5\right)\left(-2\right)=113$$

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

$$-5\times 4-2\left(-5\right)\left(-2\right)-5\times 3=-55$$

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

$$113-\left(-55\right)$$

Subtract -55 from 113.

$$168$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&3&4\\-5&1&-5\\-5&-2&1\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}1&-5\\-2&1\end{matrix}\right))-3det(\left(\begin{matrix}-5&-5\\-5&1\end{matrix}\right))+4det(\left(\begin{matrix}-5&1\\-5&-2\end{matrix}\right))$$

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

$$-2\left(1-\left(-2\left(-5\right)\right)\right)-3\left(-5-\left(-5\left(-5\right)\right)\right)+4\left(-5\left(-2\right)-\left(-5\right)\right)$$

Simplify.

$$-2\left(-9\right)-3\left(-30\right)+4\times 15$$

Add the terms to obtain the final result.

$$168$$

Factor

$$2^{3}\times 3\times 7$$