Solve {lll}{2}&{4}&{6}{2}&{5}&{8}{1}&{4}&{9}

$$\left| \begin{array} { l l l } { 2 } & { 4 } & { 6 } \\ { 2 } & { 5 } & { 8 } \\ { 1 } & { 4 } & { 9 } \end{array} \right|$$

Evaluate

$$4$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&4&6\\2&5&8\\1&4&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&4&6&2&4\\2&5&8&2&5\\1&4&9&1&4\end{matrix}\right)$$

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

$$2\times 5\times 9+4\times 8+6\times 2\times 4=170$$

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

$$5\times 6+4\times 8\times 2+9\times 2\times 4=166$$

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

$$170-166$$

Subtract 166 from 170.

$$4$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&4&6\\2&5&8\\1&4&9\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}5&8\\4&9\end{matrix}\right))-4det(\left(\begin{matrix}2&8\\1&9\end{matrix}\right))+6det(\left(\begin{matrix}2&5\\1&4\end{matrix}\right))$$

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

$$2\left(5\times 9-4\times 8\right)-4\left(2\times 9-8\right)+6\left(2\times 4-5\right)$$

Simplify.

$$2\times 13-4\times 10+6\times 3$$

Add the terms to obtain the final result.

$$4$$

Factor

$$2^{2}$$