Solve {lll}{3}&{2}&{2}{0}&{2}&{2}{0}&{0}&{-1}=

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

Calculate Determinant

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 2\left(-1\right)=-6$$

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

$$\text{true}$$

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

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-6$$

Evaluate

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