Solve {lll}{3}&{5}&{1}{x}&{0}&{1}{-4}&{-6}&{1}

$$\left| \begin{array} { l l l } { 3 } & { 5 } & { 1 } \\ { x } & { 0 } & { 1 } \\ { - 4 } & { - 6 } & { 1 } \end{array} \right|$$

Evaluate

$$-11x-2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&5&1\\x&0&1\\-4&-6&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&5&1&3&5\\x&0&1&x&0\\-4&-6&1&-4&-6\end{matrix}\right)$$

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

$$5\left(-4\right)+x\left(-6\right)=-6x-20$$

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

$$-6\times 3+x\times 5=5x-18$$

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

$$-6x-20-\left(5x-18\right)$$

Subtract -18+5x from -20-6x.

$$-11x-2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&5&1\\x&0&1\\-4&-6&1\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}0&1\\-6&1\end{matrix}\right))-5det(\left(\begin{matrix}x&1\\-4&1\end{matrix}\right))+det(\left(\begin{matrix}x&0\\-4&-6\end{matrix}\right))$$

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

$$3\left(-\left(-6\right)\right)-5\left(x-\left(-4\right)\right)+x\left(-6\right)$$

Simplify.

$$3\times 6-5\left(x+4\right)-6x$$

Add the terms to obtain the final result.

$$-11x-2$$

Integrate w.r.t. x

$$-\frac{11x^{2}}{2}-2x+С$$