Solve {rrr}{1}&{1}&{-2}{1}&{1}&{0}{-4}&{6}&{-2}

$$\left( \begin{array} { r r r } { 1 } & { 1 } & { - 2 } \\ { 1 } & { 1 } & { 0 } \\ { - 4 } & { 6 } & { - 2 } \end{array} \right)$$

Calculate Determinant

$$-20$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&-2\\1&1&0\\-4&6&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&-2&1&1\\1&1&0&1&1\\-4&6&-2&-4&6\end{matrix}\right)$$

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

$$-2-2\times 6=-14$$

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

$$-4\left(-2\right)-2=6$$

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

$$-14-6$$

Subtract 6 from -14.

$$-20$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&-2\\1&1&0\\-4&6&-2\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}1&0\\6&-2\end{matrix}\right))-det(\left(\begin{matrix}1&0\\-4&-2\end{matrix}\right))-2det(\left(\begin{matrix}1&1\\-4&6\end{matrix}\right))$$

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

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

Simplify.

$$-2-\left(-2\right)-2\times 10$$

Add the terms to obtain the final result.

$$-20$$

Evaluate

$$\left(\begin{matrix}1&1&-2\\1&1&0\\-4&6&-2\end{matrix}\right)$$