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

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

Evaluate

$$18$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2+2\left(-2\right)\times 4-2\times 4\times 2=-30$$

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

$$-12-\left(-30\right)$$

Subtract -30 from -12.

$$18$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2\left(-6\right)+6$$

Add the terms to obtain the final result.

$$18$$

Factor

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