Solve {lll}{1}&{2}&{3}{4}&{6}&{7}{5}&{8}&{9}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 3 } \\ { 4 } & { 6 } & { 7 } \\ { 5 } & { 8 } & { 9 } \end{array} \right|$$

Evaluate

$$2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\4&6&7\\5&8&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\4&6&7&4&6\\5&8&9&5&8\end{matrix}\right)$$

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

$$6\times 9+2\times 7\times 5+3\times 4\times 8=220$$

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

$$5\times 6\times 3+8\times 7+9\times 4\times 2=218$$

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

$$220-218$$

Subtract 218 from 220.

$$2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\4&6&7\\5&8&9\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}6&7\\8&9\end{matrix}\right))-2det(\left(\begin{matrix}4&7\\5&9\end{matrix}\right))+3det(\left(\begin{matrix}4&6\\5&8\end{matrix}\right))$$

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

$$6\times 9-8\times 7-2\left(4\times 9-5\times 7\right)+3\left(4\times 8-5\times 6\right)$$

Simplify.

$$-2-2+3\times 2$$

Add the terms to obtain the final result.

$$2$$

Factor

$$2$$