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

$$\left( \begin{array} { c c c } { 1 } & { 2 } & { 2 } \\ { 2 } & { 1 } & { - 2 } \\ { 2 } & { - 2 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$-27$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$1+2\left(-2\right)\times 2+2\times 2\left(-2\right)=-15$$

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

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

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

$$-15-12$$

Subtract 12 from -15.

$$-27$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$1-\left(-2\left(-2\right)\right)-2\left(2-2\left(-2\right)\right)+2\left(2\left(-2\right)-2\right)$$

Simplify.

$$-3-2\times 6+2\left(-6\right)$$

Add the terms to obtain the final result.

$$-27$$

Evaluate

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