Solve {lll}{2}&{2}&{3}{4}&{5}&{6}{5}&{7}&{6}

$$\left| \begin{array} { l l l } { 2 } & { 2 } & { 3 } \\ { 4 } & { 5 } & { 6 } \\ { 5 } & { 7 } & { 6 } \end{array} \right|$$

Evaluate

$$-3$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&2&3\\4&5&6\\5&7&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&2&3&2&2\\4&5&6&4&5\\5&7&6&5&7\end{matrix}\right)$$

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

$$2\times 5\times 6+2\times 6\times 5+3\times 4\times 7=204$$

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

$$5\times 5\times 3+7\times 6\times 2+6\times 4\times 2=207$$

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

$$204-207$$

Subtract 207 from 204.

$$-3$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&2&3\\4&5&6\\5&7&6\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}5&6\\7&6\end{matrix}\right))-2det(\left(\begin{matrix}4&6\\5&6\end{matrix}\right))+3det(\left(\begin{matrix}4&5\\5&7\end{matrix}\right))$$

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

$$2\left(5\times 6-7\times 6\right)-2\left(4\times 6-5\times 6\right)+3\left(4\times 7-5\times 5\right)$$

Simplify.

$$2\left(-12\right)-2\left(-6\right)+3\times 3$$

Add the terms to obtain the final result.

$$-3$$

Factor

$$-3$$