Solve {ccc}{2}&{1}&{4}{-1}&{2}&{4}{3}&{1}&{-1}

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

Evaluate

$$-29$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$3\times 2\times 4+4\times 2-\left(-1\right)=33$$

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

$$4-33$$

Subtract 33 from 4.

$$-29$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-6\right)-\left(-11\right)+4\left(-7\right)$$

Add the terms to obtain the final result.

$$-29$$

Factor

$$-29$$