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

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

Evaluate

$$79$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$6\times 7\times 2+5\times 6+9\times 8\times 3=330$$

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

$$7\times 9+3\times 6\times 6+2\times 8\times 5=251$$

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

$$330-251$$

Subtract 251 from 330.

$$79$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$6\left(-4\right)-5\times 10+9\times 17$$

Add the terms to obtain the final result.

$$79$$

Factor

$$79$$