Solve {lll}{6}&{8}&{9}{2}&{5}&{4}{3}&{1}&{2}

$$\left| \begin{array} { l l l } { 6 } & { 8 } & { 9 } \\ { 2 } & { 5 } & { 4 } \\ { 3 } & { 1 } & { 2 } \end{array} \right|$$

Evaluate

$$-17$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}6&8&9\\2&5&4\\3&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}6&8&9&6&8\\2&5&4&2&5\\3&1&2&3&1\end{matrix}\right)$$

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

$$6\times 5\times 2+8\times 4\times 3+9\times 2=174$$

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

$$3\times 5\times 9+4\times 6+2\times 2\times 8=191$$

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

$$174-191$$

Subtract 191 from 174.

$$-17$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}6&8&9\\2&5&4\\3&1&2\end{matrix}\right))$$

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.

$$6det(\left(\begin{matrix}5&4\\1&2\end{matrix}\right))-8det(\left(\begin{matrix}2&4\\3&2\end{matrix}\right))+9det(\left(\begin{matrix}2&5\\3&1\end{matrix}\right))$$

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

$$6\left(5\times 2-4\right)-8\left(2\times 2-3\times 4\right)+9\left(2-3\times 5\right)$$

Simplify.

$$6\times 6-8\left(-8\right)+9\left(-13\right)$$

Add the terms to obtain the final result.

$$-17$$

Factor

$$-17$$