Solve {lll}{1}&{1}&{-1}{1}&{1}&{1}{1}&{-1}&{1}

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

Calculate Determinant

$$4$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-1-1+1=-1$$

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

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

Subtract -1 from 3.

$$4$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$4$$

Evaluate

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