Solve {ccc}{0}&{-1}&{1}{2}&{4}&{1}{2}&{-4}&{1}

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

Evaluate

$$-16$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2+2\left(-4\right)=-10$$

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

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

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

$$-10-6$$

Subtract 6 from -10.

$$-16$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-16$$

Factor

$$-16$$