Solve {ccc}{18}&{4}&{5}{13}&{-1}&{8}{20}&{-2}&{7}

$$\left| \begin{array} { c c c } { 18 } & { 4 } & { 5 } \\ { 13 } & { - 1 } & { 8 } \\ { 20 } & { - 2 } & { 7 } \end{array} \right|$$

Evaluate

$$408$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}18&4&5\\13&-1&8\\20&-2&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}18&4&5&18&4\\13&-1&8&13&-1\\20&-2&7&20&-2\end{matrix}\right)$$

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

$$18\left(-1\right)\times 7+4\times 8\times 20+5\times 13\left(-2\right)=384$$

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

$$20\left(-1\right)\times 5-2\times 8\times 18+7\times 13\times 4=-24$$

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

$$384-\left(-24\right)$$

Subtract -24 from 384.

$$408$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}18&4&5\\13&-1&8\\20&-2&7\end{matrix}\right))$$

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.

$$18det(\left(\begin{matrix}-1&8\\-2&7\end{matrix}\right))-4det(\left(\begin{matrix}13&8\\20&7\end{matrix}\right))+5det(\left(\begin{matrix}13&-1\\20&-2\end{matrix}\right))$$

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

$$18\left(-7-\left(-2\times 8\right)\right)-4\left(13\times 7-20\times 8\right)+5\left(13\left(-2\right)-20\left(-1\right)\right)$$

Simplify.

$$18\times 9-4\left(-69\right)+5\left(-6\right)$$

Add the terms to obtain the final result.

$$408$$

Factor

$$2^{3}\times 3\times 17$$