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

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

Evaluate

$$-59$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 8\times 10+4\times 3\times 8+6\times 5\times 9=606$$

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

$$8\times 8\times 6+9\times 3\times 3+10\times 5\times 4=665$$

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

$$606-665$$

Subtract 665 from 606.

$$-59$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 53-4\times 26+6\left(-19\right)$$

Add the terms to obtain the final result.

$$-59$$

Factor

$$-59$$