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

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

Evaluate

$$103$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$17\times 6-3\left(-11\right)\left(-1\right)=69$$

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

$$-17\times 2=-34$$

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

$$69-\left(-34\right)$$

Subtract -34 from 69.

$$103$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$102-\left(-3\left(-11\right)\right)+2\times 17$$

Add the terms to obtain the final result.

$$103$$

Factor

$$103$$