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

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

Evaluate

$$78$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 7+6\times 5\times 2+9\times 8\times 3=297$$

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

$$2\times 7\times 9+3\times 5\times 3+8\times 6=219$$

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

$$297-219$$

Subtract 219 from 297.

$$78$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$78$$

Factor

$$2\times 3\times 13$$