Solve {lll}{x}&{y}&{z}{2}&{-1}&{1}{1}&{-1}&{1}

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

Evaluate

$$-y-z$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-z-x+2y=-x+2y-z$$

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

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

Subtract -z-x+2y from -x+y-2z.

$$-y-z$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-y+z\left(-1\right)$$

Add the terms to obtain the final result.

$$-y-z$$

Integrate w.r.t. y

$$-yz-\frac{y^{2}}{2}+С$$