Solve {ccc}{8}&{-1}&{9}{3}&{1}&{8}{11}&{0}&{17}

$$\left( \begin{array} { c c c } { 8 } & { - 1 } & { 9 } \\ { 3 } & { 1 } & { 8 } \\ { 11 } & { 0 } & { 17 } \end{array} \right)$$

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}8&-1&9\\3&1&8\\11&0&17\end{matrix}\right))$$

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

$$\left(\begin{matrix}8&-1&9&8&-1\\3&1&8&3&1\\11&0&17&11&0\end{matrix}\right)$$

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

$$8\times 17-8\times 11=48$$

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

$$11\times 9+17\times 3\left(-1\right)=48$$

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

$$48-48$$

Subtract 48 from 48.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}8&-1&9\\3&1&8\\11&0&17\end{matrix}\right))$$

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.

$$8det(\left(\begin{matrix}1&8\\0&17\end{matrix}\right))-\left(-det(\left(\begin{matrix}3&8\\11&17\end{matrix}\right))\right)+9det(\left(\begin{matrix}3&1\\11&0\end{matrix}\right))$$

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

$$8\times 17-\left(-\left(3\times 17-11\times 8\right)\right)+9\left(-11\right)$$

Simplify.

$$8\times 17-\left(-\left(-37\right)\right)+9\left(-11\right)$$

Add the terms to obtain the final result.

$$0$$

Evaluate

$$\left(\begin{matrix}8&-1&9\\3&1&8\\11&0&17\end{matrix}\right)$$