Solve {lll}{1}&{1}&{1}{z_{1}}&{z_{2}}&{z_{3}}{w_{1}}&{w_{2}}&{w_{3}}

$$\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { z _ { 1 } } & { z _ { 2 } } & { z _ { 3 } } \\ { w _ { 1 } } & { w _ { 2 } } & { w _ { 3 } } \end{array} \right|$$

Evaluate

$$w_{2}z_{1}-w_{3}z_{1}+w_{3}z_{2}-w_{1}z_{2}+w_{1}z_{3}-w_{2}z_{3}$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\z_{1}&z_{2}&z_{3}\\w_{1}&w_{2}&w_{3}\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\z_{1}&z_{2}&z_{3}&z_{1}&z_{2}\\w_{1}&w_{2}&w_{3}&w_{1}&w_{2}\end{matrix}\right)$$

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

$$z_{2}w_{3}+z_{3}w_{1}+z_{1}w_{2}=w_{2}z_{1}+w_{3}z_{2}+w_{1}z_{3}$$

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

$$w_{1}z_{2}+w_{2}z_{3}+w_{3}z_{1}=w_{3}z_{1}+w_{1}z_{2}+w_{2}z_{3}$$

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

$$w_{2}z_{1}+w_{3}z_{2}+w_{1}z_{3}-\left(w_{3}z_{1}+w_{1}z_{2}+w_{2}z_{3}\right)$$

Subtract w_{1}z_{2}+w_{2}z_{3}+w_{3}z_{1} from z_{2}w_{3}+z_{3}w_{1}+z_{1}w_{2}.

$$w_{2}z_{1}-w_{3}z_{1}+w_{3}z_{2}-w_{1}z_{2}+w_{1}z_{3}-w_{2}z_{3}$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\z_{1}&z_{2}&z_{3}\\w_{1}&w_{2}&w_{3}\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}z_{2}&z_{3}\\w_{2}&w_{3}\end{matrix}\right))-det(\left(\begin{matrix}z_{1}&z_{3}\\w_{1}&w_{3}\end{matrix}\right))+det(\left(\begin{matrix}z_{1}&z_{2}\\w_{1}&w_{2}\end{matrix}\right))$$

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

$$z_{2}w_{3}-w_{2}z_{3}-\left(z_{1}w_{3}-w_{1}z_{3}\right)+z_{1}w_{2}-w_{1}z_{2}$$

Simplify.

$$w_{3}z_{2}-w_{2}z_{3}-\left(w_{3}z_{1}-w_{1}z_{3}\right)+w_{2}z_{1}-w_{1}z_{2}$$

Add the terms to obtain the final result.

$$w_{2}z_{1}-w_{3}z_{1}+w_{3}z_{2}-w_{1}z_{2}+w_{1}z_{3}-w_{2}z_{3}$$

Integrate w.r.t. z_2

$$\frac{w_{3}z_{2}^{2}}{2}-\frac{w_{1}z_{2}^{2}}{2}+w_{1}z_{2}z_{3}+w_{2}z_{1}z_{2}-w_{2}z_{2}z_{3}-w_{3}z_{1}z_{2}+С$$