Solve {lll}{5}&{3}&{6}{2}&{1}&{3}{4}&{4}&{8}

$$\left| \begin{array} { l l l } { 5 } & { 3 } & { 6 } \\ { 2 } & { 1 } & { 3 } \\ { 4 } & { 4 } & { 8 } \end{array} \right|$$

Evaluate

$$-8$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&3&6\\2&1&3\\4&4&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&3&6&5&3\\2&1&3&2&1\\4&4&8&4&4\end{matrix}\right)$$

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

$$5\times 8+3\times 3\times 4+6\times 2\times 4=124$$

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

$$4\times 6+4\times 3\times 5+8\times 2\times 3=132$$

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

$$124-132$$

Subtract 132 from 124.

$$-8$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&3&6\\2&1&3\\4&4&8\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}1&3\\4&8\end{matrix}\right))-3det(\left(\begin{matrix}2&3\\4&8\end{matrix}\right))+6det(\left(\begin{matrix}2&1\\4&4\end{matrix}\right))$$

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

$$5\left(8-4\times 3\right)-3\left(2\times 8-4\times 3\right)+6\left(2\times 4-4\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$-8$$

Factor

$$-8$$