Solve {ccc}{13}&{11}&{1}{5}&{17}&{0}{1}&{6}&{-2}

$$\left| \begin{array} { c c c } { 13 } & { 11 } & { 1 } \\ { 5 } & { 17 } & { 0 } \\ { 1 } & { 6 } & { - 2 } \end{array} \right|$$

Evaluate

$$-319$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}13&11&1\\5&17&0\\1&6&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}13&11&1&13&11\\5&17&0&5&17\\1&6&-2&1&6\end{matrix}\right)$$

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

$$13\times 17\left(-2\right)+5\times 6=-412$$

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

$$17-2\times 5\times 11=-93$$

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

$$-412-\left(-93\right)$$

Subtract -93 from -412.

$$-319$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}13&11&1\\5&17&0\\1&6&-2\end{matrix}\right))$$

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.

$$13det(\left(\begin{matrix}17&0\\6&-2\end{matrix}\right))-11det(\left(\begin{matrix}5&0\\1&-2\end{matrix}\right))+det(\left(\begin{matrix}5&17\\1&6\end{matrix}\right))$$

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

$$13\times 17\left(-2\right)-11\times 5\left(-2\right)+5\times 6-17$$

Simplify.

$$13\left(-34\right)-11\left(-10\right)+13$$

Add the terms to obtain the final result.

$$-319$$

Factor

$$-319$$