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

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

Evaluate

$$-57$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 7\times 9+7\times 2\times 6+2\times 4\times 3=297$$

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

$$6\times 7\times 2+3\times 2\times 3+9\times 4\times 7=354$$

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

$$297-354$$

Subtract 354 from 297.

$$-57$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 57-7\times 24+2\left(-30\right)$$

Add the terms to obtain the final result.

$$-57$$

Factor

$$-57$$