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

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

Evaluate

$$172-22x$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2x+4\times 3\times 4+5\times 6\times 6=2x+228$$

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

$$4\times 5+6\times 3\times 2+x\times 6\times 4=24x+56$$

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

$$2x+228-\left(24x+56\right)$$

Subtract 56+24x from 228+2x.

$$172-22x$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(x-18\right)-4\left(6x-12\right)+5\times 32$$

Add the terms to obtain the final result.

$$172-22x$$

Integrate w.r.t. x

$$С+172x-11x^{2}$$