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

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

Evaluate

$$-7$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 7\times 2+6\times 4\times 5+7\times 6=232$$

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

$$5\times 7+6\times 4\times 5+2\times 7\times 6=239$$

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

$$232-239$$

Subtract 239 from 232.

$$-7$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$5\left(-10\right)-6\left(-6\right)+7$$

Add the terms to obtain the final result.

$$-7$$

Factor

$$-7$$