Solve {ccc}{1}&{0}&{6}{-3}&{4}&{30}{-1}&{-2}&{8}

$$\left| \begin{array} { c c c } { 1 } & { 0 } & { 6 } \\ { - 3 } & { 4 } & { 30 } \\ { - 1 } & { - 2 } & { 8 } \end{array} \right|$$

Evaluate

$$152$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&6\\-3&4&30\\-1&-2&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&6&1&0\\-3&4&30&-3&4\\-1&-2&8&-1&-2\end{matrix}\right)$$

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

$$4\times 8+6\left(-3\right)\left(-2\right)=68$$

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

$$-4\times 6-2\times 30=-84$$

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

$$68-\left(-84\right)$$

Subtract -84 from 68.

$$152$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&6\\-3&4&30\\-1&-2&8\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&30\\-2&8\end{matrix}\right))+6det(\left(\begin{matrix}-3&4\\-1&-2\end{matrix}\right))$$

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

$$4\times 8-\left(-2\times 30\right)+6\left(-3\left(-2\right)-\left(-4\right)\right)$$

Simplify.

$$92+6\times 10$$

Add the terms to obtain the final result.

$$152$$

Factor

$$2^{3}\times 19$$