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

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

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

Subtract -6 from -6.

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Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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Factor

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