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

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

Evaluate

$$-8$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2\times 4-4=4$$

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

$$-4-4$$

Subtract 4 from -4.

$$-8$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$8-16$$

Add the terms to obtain the final result.

$$-8$$

Factor

$$-8$$