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

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

Evaluate

$$-16$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 4+4\times 6+5\times 2\times 3=66$$

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

$$4\times 5+3\times 6\times 3+2\times 4=82$$

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

$$66-82$$

Subtract 82 from 66.

$$-16$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-14\right)-4\left(-4\right)+5\times 2$$

Add the terms to obtain the final result.

$$-16$$

Factor

$$-16$$