Solve {lll}{2}&{4}&{6}{7}&{5}&{3}{1}&{7}&{4}

$$\left| \begin{array} { l l l } { 2 } & { 4 } & { 6 } \\ { 7 } & { 5 } & { 3 } \\ { 1 } & { 7 } & { 4 } \end{array} \right|$$

Evaluate

$$162$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&4&6\\7&5&3\\1&7&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&4&6&2&4\\7&5&3&7&5\\1&7&4&1&7\end{matrix}\right)$$

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

$$2\times 5\times 4+4\times 3+6\times 7\times 7=346$$

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

$$5\times 6+7\times 3\times 2+4\times 7\times 4=184$$

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

$$346-184$$

Subtract 184 from 346.

$$162$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&4&6\\7&5&3\\1&7&4\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}5&3\\7&4\end{matrix}\right))-4det(\left(\begin{matrix}7&3\\1&4\end{matrix}\right))+6det(\left(\begin{matrix}7&5\\1&7\end{matrix}\right))$$

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

$$2\left(5\times 4-7\times 3\right)-4\left(7\times 4-3\right)+6\left(7\times 7-5\right)$$

Simplify.

$$2\left(-1\right)-4\times 25+6\times 44$$

Add the terms to obtain the final result.

$$162$$

Factor

$$2\times 3^{4}$$