Solve {lll}{3}&{2}&{-2}{1}&{1}&{1}{2}&{-3}&{3}

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

Calculate Determinant

$$26$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3+2\times 2-2\left(-3\right)=19$$

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

$$2\left(-2\right)-3\times 3+3\times 2=-7$$

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

$$19-\left(-7\right)$$

Subtract -7 from 19.

$$26$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 6-2-2\left(-5\right)$$

Add the terms to obtain the final result.

$$26$$

Evaluate

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