Solve {ccc}{1}&{2}&{2}{6}&{4}&{-4}{-2}&{1/2}&{5}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { 2 } \\ { 6 } & { 4 } & { - 4 } \\ { - 2 } & { \frac { 1 } { 2 } } & { 5 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&2\\6&4&-4\\-2&\frac{1}{2}&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&2&1&2\\6&4&-4&6&4\\-2&\frac{1}{2}&5&-2&\frac{1}{2}\end{matrix}\right)$$

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

$$4\times 5+2\left(-4\right)\left(-2\right)+2\times 6\times \frac{1}{2}=42$$

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

$$-2\times 4\times 2+\frac{1}{2}\left(-4\right)+5\times 6\times 2=42$$

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

$$42-42$$

Subtract 42 from 42.

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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).

$$det(\left(\begin{matrix}1&2&2\\6&4&-4\\-2&\frac{1}{2}&5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&-4\\\frac{1}{2}&5\end{matrix}\right))-2det(\left(\begin{matrix}6&-4\\-2&5\end{matrix}\right))+2det(\left(\begin{matrix}6&4\\-2&\frac{1}{2}\end{matrix}\right))$$

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

$$4\times 5-\frac{1}{2}\left(-4\right)-2\left(6\times 5-\left(-2\left(-4\right)\right)\right)+2\left(6\times \frac{1}{2}-\left(-2\times 4\right)\right)$$

Simplify.

$$22-2\times 22+2\times 11$$

Add the terms to obtain the final result.

$$0$$

Factor

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