Solve {lll}{3}&{0}&{3}{1}&{2}&{8}{5}&{7}&{9}

$$\left| \begin{array} { l l l } { 3 } & { 0 } & { 3 } \\ { 1 } & { 2 } & { 8 } \\ { 5 } & { 7 } & { 9 } \end{array} \right|$$

Evaluate

$$-123$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&0&3\\1&2&8\\5&7&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&0&3&3&0\\1&2&8&1&2\\5&7&9&5&7\end{matrix}\right)$$

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

$$3\times 2\times 9+3\times 7=75$$

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

$$5\times 2\times 3+7\times 8\times 3=198$$

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

$$75-198$$

Subtract 198 from 75.

$$-123$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&0&3\\1&2&8\\5&7&9\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}2&8\\7&9\end{matrix}\right))+3det(\left(\begin{matrix}1&2\\5&7\end{matrix}\right))$$

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

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

Simplify.

$$3\left(-38\right)+3\left(-3\right)$$

Add the terms to obtain the final result.

$$-123$$

Factor

$$-123$$