Solve {lll}{2}&{3}&{4}{4}&{9}&{16}{2}&{2}&{2}=

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

Evaluate

$$4$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 9\times 2+3\times 16\times 2+4\times 4\times 2=164$$

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

$$2\times 9\times 4+2\times 16\times 2+2\times 4\times 3=160$$

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

$$164-160$$

Subtract 160 from 164.

$$4$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-14\right)-3\left(-24\right)+4\left(-10\right)$$

Add the terms to obtain the final result.

$$4$$

Factor

$$2^{2}$$