Solve {lll}{3}&{7}&{2}{6}&{8}&{3}{9}&{4}&{1}

$$\left| \begin{array} { l l l } { 3 } & { 7 } & { 2 } \\ { 6 } & { 8 } & { 3 } \\ { 9 } & { 4 } & { 1 } \end{array} \right|$$

Evaluate

$$39$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&7&2\\6&8&3\\9&4&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&7&2&3&7\\6&8&3&6&8\\9&4&1&9&4\end{matrix}\right)$$

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

$$3\times 8+7\times 3\times 9+2\times 6\times 4=261$$

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

$$9\times 8\times 2+4\times 3\times 3+6\times 7=222$$

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

$$261-222$$

Subtract 222 from 261.

$$39$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&7&2\\6&8&3\\9&4&1\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}8&3\\4&1\end{matrix}\right))-7det(\left(\begin{matrix}6&3\\9&1\end{matrix}\right))+2det(\left(\begin{matrix}6&8\\9&4\end{matrix}\right))$$

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

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

Simplify.

$$3\left(-4\right)-7\left(-21\right)+2\left(-48\right)$$

Add the terms to obtain the final result.

$$39$$

Factor

$$3\times 13$$