Solve {ccc}{x}&{y}&{z}{5}&{2}&{1}{1}&{-4}&{2}

$$\left| \begin{array} { c c c } { x } & { y } & { z } \\ { 5 } & { 2 } & { 1 } \\ { 1 } & { - 4 } & { 2 } \end{array} \right|$$

Evaluate

$$8x-9y-22z$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}x&y&z\\5&2&1\\1&-4&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}x&y&z&x&y\\5&2&1&5&2\\1&-4&2&1&-4\end{matrix}\right)$$

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

$$x\times 2\times 2+y+z\times 5\left(-4\right)=4x+y-20z$$

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

$$2z-4x+2\times 5y=2z+10y-4x$$

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

$$4x+y-20z-\left(2z+10y-4x\right)$$

Subtract 2z-4x+10y from 4x+y-20z.

$$8x-9y-22z$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}x&y&z\\5&2&1\\1&-4&2\end{matrix}\right))$$

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.

$$xdet(\left(\begin{matrix}2&1\\-4&2\end{matrix}\right))-ydet(\left(\begin{matrix}5&1\\1&2\end{matrix}\right))+zdet(\left(\begin{matrix}5&2\\1&-4\end{matrix}\right))$$

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

$$x\left(2\times 2-\left(-4\right)\right)-y\left(5\times 2-1\right)+z\left(5\left(-4\right)-2\right)$$

Simplify.

$$x\times 8-y\times 9+z\left(-22\right)$$

Add the terms to obtain the final result.

$$8x-9y-22z$$

Integrate w.r.t. x

$$4x^{2}-9xy-22xz+С$$