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

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

Calculate Determinant

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-1+2\times 2\times 2-\left(-1\right)=8$$

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

$$2-8$$

Subtract 8 from 2.

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-5\right)-\left(-1\right)-\left(-3\right)$$

Add the terms to obtain the final result.

$$-6$$

Evaluate

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