Solve {ccc}{2}&{3}&{5}{4}&{9}&{25}{16}&{27}&{125}

$$\left| \begin{array} { c c c } { 2 } & { 3 } & { 5 } \\ { 4 } & { 9 } & { 25 } \\ { 16 } & { 27 } & { 125 } \end{array} \right|$$

Evaluate

$$420$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&5\\4&9&25\\16&27&125\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&5&2&3\\4&9&25&4&9\\16&27&125&16&27\end{matrix}\right)$$

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

$$2\times 9\times 125+3\times 25\times 16+5\times 4\times 27=3990$$

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

$$16\times 9\times 5+27\times 25\times 2+125\times 4\times 3=3570$$

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

$$3990-3570$$

Subtract 3570 from 3990.

$$420$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&5\\4&9&25\\16&27&125\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}9&25\\27&125\end{matrix}\right))-3det(\left(\begin{matrix}4&25\\16&125\end{matrix}\right))+5det(\left(\begin{matrix}4&9\\16&27\end{matrix}\right))$$

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

$$2\left(9\times 125-27\times 25\right)-3\left(4\times 125-16\times 25\right)+5\left(4\times 27-16\times 9\right)$$

Simplify.

$$2\times 450-3\times 100+5\left(-36\right)$$

Add the terms to obtain the final result.

$$420$$

Factor

$$2^{2}\times 3\times 5\times 7$$