Solve {rrr}{1}&{1}&{-1}{2}&{1}&{-2}{1}&{0}&{1}^-1{rrr}{1}&{-1}&{1}{2}&{-2}&{2}{-1}&{1}&{-1}{rrr}{1}&{1}&{-1}{2}&{1}&{-2}{1}&{0}&{1}

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

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

Multiply \left(\left(\begin{matrix}1&1&-1\\2&1&-2\\1&0&1\end{matrix}\right)\right)^{-1} and \left(\begin{matrix}1&1&-1\\2&1&-2\\1&0&1\end{matrix}\right) to get 1.

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

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

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

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

$$-2\left(-1\right)-2\left(-1\right)+2=6$$

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

$$-\left(-2\right)+2-2\left(-1\right)=6$$

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

Multiply \left(\left(\begin{matrix}1&1&-1\\2&1&-2\\1&0&1\end{matrix}\right)\right)^{-1} and \left(\begin{matrix}1&1&-1\\2&1&-2\\1&0&1\end{matrix}\right) to get 1.

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

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.

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

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

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

Simplify.

$$0$$

Evaluate

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