Solve {lll}{x}&{0}&{1}{x}&{1}&{1}{x}&{x}&{x}

$$\left| \begin{array} { l l l } { x } & { 0 } & { 1 } \\ { x } & { 1 } & { 1 } \\ { x } & { x } & { x } \end{array} \right|$$

Evaluate

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

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}x&0&1\\x&1&1\\x&x&x\end{matrix}\right))$$

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

$$\left(\begin{matrix}x&0&1&x&0\\x&1&1&x&1\\x&x&x&x&x\end{matrix}\right)$$

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

$$xx+xx=2x^{2}$$

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

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

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

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

Subtract x\left(1+x\right) from 2x^{2}.

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

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}x&0&1\\x&1&1\\x&x&x\end{matrix}\right))$$

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.

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

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

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

Simplify.

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

Integrate w.r.t. x

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}+С$$