Solve {ccc}{6}&{-6}&{2}{4}&{3}&{-2}{1}&{-2}&{0}

$$\left| \begin{array} { c c c } { 6 } & { - 6 } & { 2 } \\ { 4 } & { 3 } & { - 2 } \\ { 1 } & { - 2 } & { 0 } \end{array} \right|$$

Evaluate

$$-34$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}6&-6&2\\4&3&-2\\1&-2&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}6&-6&2&6&-6\\4&3&-2&4&3\\1&-2&0&1&-2\end{matrix}\right)$$

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

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

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

$$3\times 2-2\left(-2\right)\times 6=30$$

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

$$-4-30$$

Subtract 30 from -4.

$$-34$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}6&-6&2\\4&3&-2\\1&-2&0\end{matrix}\right))$$

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.

$$6det(\left(\begin{matrix}3&-2\\-2&0\end{matrix}\right))-\left(-6det(\left(\begin{matrix}4&-2\\1&0\end{matrix}\right))\right)+2det(\left(\begin{matrix}4&3\\1&-2\end{matrix}\right))$$

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

$$6\left(-\left(-2\left(-2\right)\right)\right)-\left(-6\left(-\left(-2\right)\right)\right)+2\left(4\left(-2\right)-3\right)$$

Simplify.

$$6\left(-4\right)-\left(-6\times 2\right)+2\left(-11\right)$$

Add the terms to obtain the final result.

$$-34$$

Factor

$$-34$$