Solve {rrr}{3}&{2}&{4}{1}&{-1}&{2}{4}&{-1}&{8}

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

Evaluate

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\left(-1\right)\times 8+2\times 2\times 4+4\left(-1\right)=-12$$

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

$$4\left(-1\right)\times 4-2\times 3+8\times 2=-6$$

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

$$-12-\left(-6\right)$$

Subtract -6 from -12.

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-6$$

Factor

$$-6$$