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

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

Evaluate

$$93$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 8\times 7+2\times 7\times 9+6\times 3=312$$

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

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

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

$$312-219$$

Subtract 219 from 312.

$$93$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 35-2\left(-21\right)-54$$

Add the terms to obtain the final result.

$$93$$

Factor

$$3\times 31$$