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

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

Evaluate

$$-60$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$7\times 7+8\times 2\times 3+9\times 6\times 2=205$$

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

$$3\times 7\times 9+2\times 2\times 7+6\times 8=265$$

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

$$205-265$$

Subtract 265 from 205.

$$-60$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-60$$

Factor

$$-60$$