Solve {ccc}{13}&{6}&{5}{3}&{6}&{-4}{13}&{8}&{3}

$$\left| \begin{array} { c c c } { 13 } & { 6 } & { 5 } \\ { 3 } & { 6 } & { - 4 } \\ { 13 } & { 8 } & { 3 } \end{array} \right|$$

Evaluate

$$14$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}13&6&5\\3&6&-4\\13&8&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}13&6&5&13&6\\3&6&-4&3&6\\13&8&3&13&8\end{matrix}\right)$$

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

$$13\times 6\times 3+6\left(-4\right)\times 13+5\times 3\times 8=42$$

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

$$13\times 6\times 5+8\left(-4\right)\times 13+3\times 3\times 6=28$$

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

$$42-28$$

Subtract 28 from 42.

$$14$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}13&6&5\\3&6&-4\\13&8&3\end{matrix}\right))$$

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.

$$13det(\left(\begin{matrix}6&-4\\8&3\end{matrix}\right))-6det(\left(\begin{matrix}3&-4\\13&3\end{matrix}\right))+5det(\left(\begin{matrix}3&6\\13&8\end{matrix}\right))$$

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

$$13\left(6\times 3-8\left(-4\right)\right)-6\left(3\times 3-13\left(-4\right)\right)+5\left(3\times 8-13\times 6\right)$$

Simplify.

$$13\times 50-6\times 61+5\left(-54\right)$$

Add the terms to obtain the final result.

$$14$$

Factor

$$2\times 7$$