Solve {rrr}{11}&{-2}&{1}{17}&{3}&{0}{1}&{-2}&{6}

$$\left| \begin{array} { r r r } { 11 } & { - 2 } & { 1 } \\ { 17 } & { 3 } & { 0 } \\ { 1 } & { - 2 } & { 6 } \end{array} \right|$$

Evaluate

$$365$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}11&-2&1\\17&3&0\\1&-2&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}11&-2&1&11&-2\\17&3&0&17&3\\1&-2&6&1&-2\end{matrix}\right)$$

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

$$11\times 3\times 6+17\left(-2\right)=164$$

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

$$3+6\times 17\left(-2\right)=-201$$

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

$$164-\left(-201\right)$$

Subtract -201 from 164.

$$365$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}11&-2&1\\17&3&0\\1&-2&6\end{matrix}\right))$$

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.

$$11det(\left(\begin{matrix}3&0\\-2&6\end{matrix}\right))-\left(-2det(\left(\begin{matrix}17&0\\1&6\end{matrix}\right))\right)+det(\left(\begin{matrix}17&3\\1&-2\end{matrix}\right))$$

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

$$11\times 3\times 6-\left(-2\times 17\times 6\right)+17\left(-2\right)-3$$

Simplify.

$$11\times 18-\left(-2\times 102\right)-37$$

Add the terms to obtain the final result.

$$365$$

Factor

$$5\times 73$$