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

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

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3+3\left(-1\right)=6$$

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

$$2\times 3=6$$

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

$$6-6$$

Subtract 6 from 6.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-3\left(-3\right)+3\left(-3\right)$$

Add the terms to obtain the final result.

$$0$$

Evaluate

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