Solve {ccc}{1}&{3}&{-3}{0}&{1}&{-1}{-2}&{-8}&{9}

$$\left( \begin{array} { c c c } { 1 } & { 3 } & { - 3 } \\ { 0 } & { 1 } & { - 1 } \\ { - 2 } & { - 8 } & { 9 } \end{array} \right)$$

Calculate Determinant

$$1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&-3\\0&1&-1\\-2&-8&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&-3&1&3\\0&1&-1&0&1\\-2&-8&9&-2&-8\end{matrix}\right)$$

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

$$9+3\left(-1\right)\left(-2\right)=15$$

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

$$-2\left(-3\right)-8\left(-1\right)=14$$

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

$$15-14$$

Subtract 14 from 15.

$$1$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&-3\\0&1&-1\\-2&-8&9\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}1&-1\\-8&9\end{matrix}\right))-3det(\left(\begin{matrix}0&-1\\-2&9\end{matrix}\right))-3det(\left(\begin{matrix}0&1\\-2&-8\end{matrix}\right))$$

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

$$9-\left(-8\left(-1\right)\right)-3\left(-\left(-2\left(-1\right)\right)\right)-3\left(-\left(-2\right)\right)$$

Simplify.

$$1-3\left(-2\right)-3\times 2$$

Add the terms to obtain the final result.

$$1$$

Evaluate

$$\left(\begin{matrix}1&3&-3\\0&1&-1\\-2&-8&9\end{matrix}\right)$$