Solve {lll}{x}&{1}&{2}{3}&{4}&{5}{6}&{7}&{8}=

$$\left| \begin{array} { l l l } { x } & { 1 } & { 2 } \\ { 3 } & { 4 } & { 5 } \\ { 6 } & { 7 } & { 8 } \end{array} \right| =$$

Evaluate

$$-3x$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}x&1&2\\3&4&5\\6&7&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}x&1&2&x&1\\3&4&5&3&4\\6&7&8&6&7\end{matrix}\right)$$

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

$$x\times 4\times 8+5\times 6+2\times 3\times 7=32x+72$$

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

$$6\times 4\times 2+7\times 5x+8\times 3=35x+72$$

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

$$32x+72-\left(35x+72\right)$$

Subtract 72+35x from 32x+72.

$$-3x$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}x&1&2\\3&4&5\\6&7&8\end{matrix}\right))$$

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.

$$xdet(\left(\begin{matrix}4&5\\7&8\end{matrix}\right))-det(\left(\begin{matrix}3&5\\6&8\end{matrix}\right))+2det(\left(\begin{matrix}3&4\\6&7\end{matrix}\right))$$

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

$$x\left(4\times 8-7\times 5\right)-\left(3\times 8-6\times 5\right)+2\left(3\times 7-6\times 4\right)$$

Simplify.

$$x\left(-3\right)-\left(-6\right)+2\left(-3\right)$$

Add the terms to obtain the final result.

$$-3x$$

Integrate w.r.t. x

$$-\frac{3x^{2}}{2}+С$$