Solve {ccc}{2}&{3}&{4}{1}&{1}&{1}{4}&{9}&{16}=

$$\left| \begin{array} { c c c } { 2 } & { 3 } & { 4 } \\ { 1 } & { 1 } & { 1 } \\ { 4 } & { 9 } & { 16 } \end{array} \right| =$$

Evaluate

$$-2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&4\\1&1&1\\4&9&16\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&4&2&3\\1&1&1&1&1\\4&9&16&4&9\end{matrix}\right)$$

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

$$2\times 16+3\times 4+4\times 9=80$$

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

$$4\times 4+9\times 2+16\times 3=82$$

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

$$80-82$$

Subtract 82 from 80.

$$-2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&4\\1&1&1\\4&9&16\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&1\\9&16\end{matrix}\right))-3det(\left(\begin{matrix}1&1\\4&16\end{matrix}\right))+4det(\left(\begin{matrix}1&1\\4&9\end{matrix}\right))$$

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

$$2\left(16-9\right)-3\left(16-4\right)+4\left(9-4\right)$$

Simplify.

$$2\times 7-3\times 12+4\times 5$$

Add the terms to obtain the final result.

$$-2$$

Factor

$$-2$$