Solve {lll}{3}&{x}&{-1}{4}&{0}&{1}{x}&{5}&{-1}

$$\left| \begin{array} { l l l } { 3 } & { x } & { - 1 } \\ { 4 } & { 0 } & { 1 } \\ { x } & { 5 } & { - 1 } \end{array} \right|$$

Evaluate

$$x^{2}+4x-35$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&x&-1\\4&0&1\\x&5&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&x&-1&3&x\\4&0&1&4&0\\x&5&-1&x&5\end{matrix}\right)$$

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

$$xx-4\times 5=x^{2}-20$$

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

$$5\times 3-4x=15-4x$$

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

$$x^{2}-20-\left(15-4x\right)$$

Subtract 15-4x from x^{2}-20.

$$x^{2}+4x-35$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&x&-1\\4&0&1\\x&5&-1\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}0&1\\5&-1\end{matrix}\right))-xdet(\left(\begin{matrix}4&1\\x&-1\end{matrix}\right))-det(\left(\begin{matrix}4&0\\x&5\end{matrix}\right))$$

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

$$3\left(-5\right)-x\left(4\left(-1\right)-x\right)-4\times 5$$

Simplify.

$$3\left(-5\right)-x\left(-x-4\right)-20$$

Add the terms to obtain the final result.

$$x^{2}+4x-35$$

Integrate w.r.t. x

$$\frac{x^{3}}{3}+2x^{2}-35x+С$$