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

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

Evaluate

$$-74$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\left(-6\right)\times 4+2\left(-1\right)\left(-3\right)=-90$$

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

$$4\times 2+6\left(-1\right)\times 4=-16$$

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

$$-90-\left(-16\right)$$

Subtract -16 from -90.

$$-74$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-4\times 18+2\left(-1\right)$$

Add the terms to obtain the final result.

$$-74$$

Factor

$$-74$$