Solve {lll}{1}&{1}&{1}{2}&{4}&{8}{5}&{15}&{125}

$$\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 2 } & { 4 } & { 8 } \\ { 5 } & { 15 } & { 125 } \end{array} \right|$$

Evaluate

$$180$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\2&4&8\\5&15&125\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\2&4&8&2&4\\5&15&125&5&15\end{matrix}\right)$$

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

$$4\times 125+8\times 5+2\times 15=570$$

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

$$5\times 4+15\times 8+125\times 2=390$$

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

$$570-390$$

Subtract 390 from 570.

$$180$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\2&4&8\\5&15&125\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&8\\15&125\end{matrix}\right))-det(\left(\begin{matrix}2&8\\5&125\end{matrix}\right))+det(\left(\begin{matrix}2&4\\5&15\end{matrix}\right))$$

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

$$4\times 125-15\times 8-\left(2\times 125-5\times 8\right)+2\times 15-5\times 4$$

Simplify.

$$380-210+10$$

Add the terms to obtain the final result.

$$180$$

Factor

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