Solve {lll}{1}&{2}&{3}{3}&{4}&{5}{8}&{9}&{7}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 3 } \\ { 3 } & { 4 } & { 5 } \\ { 8 } & { 9 } & { 7 } \end{array} \right|$$

Evaluate

$$6$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\3&4&5\\8&9&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\3&4&5&3&4\\8&9&7&8&9\end{matrix}\right)$$

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

$$4\times 7+2\times 5\times 8+3\times 3\times 9=189$$

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

$$8\times 4\times 3+9\times 5+7\times 3\times 2=183$$

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

$$189-183$$

Subtract 183 from 189.

$$6$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\3&4&5\\8&9&7\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&5\\9&7\end{matrix}\right))-2det(\left(\begin{matrix}3&5\\8&7\end{matrix}\right))+3det(\left(\begin{matrix}3&4\\8&9\end{matrix}\right))$$

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

$$4\times 7-9\times 5-2\left(3\times 7-8\times 5\right)+3\left(3\times 9-8\times 4\right)$$

Simplify.

$$-17-2\left(-19\right)+3\left(-5\right)$$

Add the terms to obtain the final result.

$$6$$

Factor

$$2\times 3$$