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

$$\left| \begin{array} { l l l } { - 1 } & { x } & { x } \\ { x } & { 1 } & { x } \\ { x } & { x } & { 1 } \end{array} \right|$$

Evaluate

$$\left(x-1\right)\left(2x^{2}+x+1\right)$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-1+xxx+xxx=2x^{3}-1$$

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

$$xx+xx\left(-1\right)+xx=x^{2}$$

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

$$2x^{3}-1-x^{2}$$

Subtract x^{2} from -1+2x^{3}.

$$\left(x-1\right)\left(2x^{2}+x+1\right)$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-\left(1-xx\right)-x\left(x-xx\right)+x\left(xx-x\right)$$

Simplify.

$$-\left(1-x^{2}\right)-xx\left(1-x\right)+xx\left(x-1\right)$$

Add the terms to obtain the final result.

$$\left(x-1\right)\left(2x^{2}+x+1\right)$$

Integrate w.r.t. x

$$\frac{x^{4}}{2}-\frac{x^{3}}{3}-x+С$$