Solve {ccc}{-2}&{1}&{4}{4}&{1}&{-2}{-6}&{-1}&{7}

$$\left| \begin{array} { c c c } { - 2 } & { 1 } & { 4 } \\ { 4 } & { 1 } & { - 2 } \\ { - 6 } & { - 1 } & { 7 } \end{array} \right|$$

Evaluate

$$-18$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&1&4\\4&1&-2\\-6&-1&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&1&4&-2&1\\4&1&-2&4&1\\-6&-1&7&-6&-1\end{matrix}\right)$$

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

$$-2\times 7-2\left(-6\right)+4\times 4\left(-1\right)=-18$$

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

$$-6\times 4-\left(-2\left(-2\right)\right)+7\times 4=0$$

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

$$-18$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&1&4\\4&1&-2\\-6&-1&7\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}1&-2\\-1&7\end{matrix}\right))-det(\left(\begin{matrix}4&-2\\-6&7\end{matrix}\right))+4det(\left(\begin{matrix}4&1\\-6&-1\end{matrix}\right))$$

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

$$-2\left(7-\left(-\left(-2\right)\right)\right)-\left(4\times 7-\left(-6\left(-2\right)\right)\right)+4\left(4\left(-1\right)-\left(-6\right)\right)$$

Simplify.

$$-2\times 5-16+4\times 2$$

Add the terms to obtain the final result.

$$-18$$

Factor

$$-18$$