Solve {rrr}{x}&{y}&{z}{7}&{-6}&{1}{4}&{6}&{8}

$$\left| \begin{array} { r r r } { x } & { y } & { z } \\ { 7 } & { - 6 } & { 1 } \\ { 4 } & { 6 } & { 8 } \end{array} \right|$$

Evaluate

$$66z-52y-54x$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}x&y&z\\7&-6&1\\4&6&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}x&y&z&x&y\\7&-6&1&7&-6\\4&6&8&4&6\end{matrix}\right)$$

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

$$x\left(-6\right)\times 8+y\times 4+z\times 7\times 6=42z+4y-48x$$

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

$$4\left(-6\right)z+6x+8\times 7y=6x+56y-24z$$

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

$$42z+4y-48x-\left(6x+56y-24z\right)$$

Subtract -24z+6x+56y from -48x+4y+42z.

$$66z-52y-54x$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}x&y&z\\7&-6&1\\4&6&8\end{matrix}\right))$$

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.

$$xdet(\left(\begin{matrix}-6&1\\6&8\end{matrix}\right))-ydet(\left(\begin{matrix}7&1\\4&8\end{matrix}\right))+zdet(\left(\begin{matrix}7&-6\\4&6\end{matrix}\right))$$

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

$$x\left(-6\times 8-6\right)-y\left(7\times 8-4\right)+z\left(7\times 6-4\left(-6\right)\right)$$

Simplify.

$$x\left(-54\right)-y\times 52+z\times 66$$

Add the terms to obtain the final result.

$$66z-52y-54x$$

Integrate w.r.t. x

$$С+66xz-52xy-27x^{2}$$