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

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

Evaluate

$$-156$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$9\times 8\times 2+6\times 7\times 3+2\times 2\times 6=294$$

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

$$3\times 8\times 2+6\times 7\times 9+2\times 2\times 6=450$$

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

$$294-450$$

Subtract 450 from 294.

$$-156$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$9\left(-26\right)-6\left(-17\right)+2\left(-12\right)$$

Add the terms to obtain the final result.

$$-156$$

Factor

$$-156$$