Solve {ccc}{4}&{4}&{-2}{2}&{8}&{-4}{1}&{1}&{1}=

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

Evaluate

$$36$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$8\left(-2\right)-4\times 4+2\times 4=-24$$

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

$$12-\left(-24\right)$$

Subtract -24 from 12.

$$36$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$4\left(8-\left(-4\right)\right)-4\left(2-\left(-4\right)\right)-2\left(2-8\right)$$

Simplify.

$$4\times 12-4\times 6-2\left(-6\right)$$

Add the terms to obtain the final result.

$$36$$

Factor

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