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

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

Evaluate

$$-84$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 2\times 7+3\times 9+8\times 2=85$$

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

$$9\times 2\times 8+2\times 3\times 3+7=169$$

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

$$85-169$$

Subtract 169 from 85.

$$-84$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 8-\left(-20\right)+8\left(-16\right)$$

Add the terms to obtain the final result.

$$-84$$

Factor

$$-84$$