Solve {lll}{3}&{4}&{5}{1}&{16}&{25}{27}&{64}&{125}

$$\left| \begin{array} { l l l } { 3 } & { 4 } & { 5 } \\ { 1 } & { 16 } & { 25 } \\ { 27 } & { 64 } & { 125 } \end{array} \right|$$

Evaluate

$$1560$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&4&5\\1&16&25\\27&64&125\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&4&5&3&4\\1&16&25&1&16\\27&64&125&27&64\end{matrix}\right)$$

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

$$3\times 16\times 125+4\times 25\times 27+5\times 64=9020$$

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

$$27\times 16\times 5+64\times 25\times 3+125\times 4=7460$$

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

$$9020-7460$$

Subtract 7460 from 9020.

$$1560$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&4&5\\1&16&25\\27&64&125\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}16&25\\64&125\end{matrix}\right))-4det(\left(\begin{matrix}1&25\\27&125\end{matrix}\right))+5det(\left(\begin{matrix}1&16\\27&64\end{matrix}\right))$$

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

$$3\left(16\times 125-64\times 25\right)-4\left(125-27\times 25\right)+5\left(64-27\times 16\right)$$

Simplify.

$$3\times 400-4\left(-550\right)+5\left(-368\right)$$

Add the terms to obtain the final result.

$$1560$$

Factor

$$2^{3}\times 3\times 5\times 13$$