Solve {lll}{2}&{3}&{9}{7}&{4}&{6}{8}&{5}&{10}

$$\left| \begin{array} { l l l } { 2 } & { 3 } & { 9 } \\ { 7 } & { 4 } & { 6 } \\ { 8 } & { 5 } & { 10 } \end{array} \right|$$

Evaluate

$$-19$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&9\\7&4&6\\8&5&10\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&9&2&3\\7&4&6&7&4\\8&5&10&8&5\end{matrix}\right)$$

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

$$2\times 4\times 10+3\times 6\times 8+9\times 7\times 5=539$$

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

$$8\times 4\times 9+5\times 6\times 2+10\times 7\times 3=558$$

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

$$539-558$$

Subtract 558 from 539.

$$-19$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&9\\7&4&6\\8&5&10\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}4&6\\5&10\end{matrix}\right))-3det(\left(\begin{matrix}7&6\\8&10\end{matrix}\right))+9det(\left(\begin{matrix}7&4\\8&5\end{matrix}\right))$$

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

$$2\left(4\times 10-5\times 6\right)-3\left(7\times 10-8\times 6\right)+9\left(7\times 5-8\times 4\right)$$

Simplify.

$$2\times 10-3\times 22+9\times 3$$

Add the terms to obtain the final result.

$$-19$$

Factor

$$-19$$