Solve {lll}{1}&{2}&{3}{3}&{0}&{1}{4}&{1}&{8}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 3 } \\ { 3 } & { 0 } & { 1 } \\ { 4 } & { 1 } & { 8 } \end{array} \right|$$

Evaluate

$$-32$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\3&0&1\\4&1&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\3&0&1&3&0\\4&1&8&4&1\end{matrix}\right)$$

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

$$2\times 4+3\times 3=17$$

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

$$1+8\times 3\times 2=49$$

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

$$17-49$$

Subtract 49 from 17.

$$-32$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\3&0&1\\4&1&8\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}0&1\\1&8\end{matrix}\right))-2det(\left(\begin{matrix}3&1\\4&8\end{matrix}\right))+3det(\left(\begin{matrix}3&0\\4&1\end{matrix}\right))$$

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

$$-1-2\left(3\times 8-4\right)+3\times 3$$

Simplify.

$$-1-2\times 20+3\times 3$$

Add the terms to obtain the final result.

$$-32$$

Factor

$$-32$$