Solve {ccc}{10}&{7}&{9}{8}&{5}&{6}{3}&{2}&{7}

$$\left| \begin{array} { c c c } { 10 } & { 7 } & { 9 } \\ { 8 } & { 5 } & { 6 } \\ { 3 } & { 2 } & { 7 } \end{array} \right|$$

Evaluate

$$-27$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}10&7&9\\8&5&6\\3&2&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}10&7&9&10&7\\8&5&6&8&5\\3&2&7&3&2\end{matrix}\right)$$

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

$$10\times 5\times 7+7\times 6\times 3+9\times 8\times 2=620$$

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

$$3\times 5\times 9+2\times 6\times 10+7\times 8\times 7=647$$

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

$$620-647$$

Subtract 647 from 620.

$$-27$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}10&7&9\\8&5&6\\3&2&7\end{matrix}\right))$$

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.

$$10det(\left(\begin{matrix}5&6\\2&7\end{matrix}\right))-7det(\left(\begin{matrix}8&6\\3&7\end{matrix}\right))+9det(\left(\begin{matrix}8&5\\3&2\end{matrix}\right))$$

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

$$10\left(5\times 7-2\times 6\right)-7\left(8\times 7-3\times 6\right)+9\left(8\times 2-3\times 5\right)$$

Simplify.

$$10\times 23-7\times 38+9$$

Add the terms to obtain the final result.

$$-27$$

Factor

$$-27$$