Solve {ccc}{6}&{4}&{-2}{1}&{5}&{6}{0}&{-1}&{7}

$$\left| \begin{array} { c c c } { 6 } & { 4 } & { - 2 } \\ { 1 } & { 5 } & { 6 } \\ { 0 } & { - 1 } & { 7 } \end{array} \right|$$

Evaluate

$$220$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}6&4&-2\\1&5&6\\0&-1&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}6&4&-2&6&4\\1&5&6&1&5\\0&-1&7&0&-1\end{matrix}\right)$$

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

$$6\times 5\times 7-2\left(-1\right)=212$$

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

$$-6\times 6+7\times 4=-8$$

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

$$212-\left(-8\right)$$

Subtract -8 from 212.

$$220$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}6&4&-2\\1&5&6\\0&-1&7\end{matrix}\right))$$

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.

$$6det(\left(\begin{matrix}5&6\\-1&7\end{matrix}\right))-4det(\left(\begin{matrix}1&6\\0&7\end{matrix}\right))-2det(\left(\begin{matrix}1&5\\0&-1\end{matrix}\right))$$

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

$$6\left(5\times 7-\left(-6\right)\right)-4\times 7-2\left(-1\right)$$

Simplify.

$$6\times 41-4\times 7-2\left(-1\right)$$

Add the terms to obtain the final result.

$$220$$

Factor

$$2^{2}\times 5\times 11$$