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

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

Evaluate

$$21$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 2+2\left(-1\right)\left(-3\right)=12$$

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

$$9\left(-1\right)=-9$$

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

$$12-\left(-9\right)$$

Subtract -9 from 12.

$$21$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\left(-15\right)+2\times 3$$

Add the terms to obtain the final result.

$$21$$

Factor

$$3\times 7$$