Solve {lll}{x}&{y}&{z}{3}&{-1}&{0}{2}&{0}&{-1}

$$\left| \begin{array} { l l l } { x } & { y } & { z } \\ { 3 } & { - 1 } & { 0 } \\ { 2 } & { 0 } & { - 1 } \end{array} \right|$$

Evaluate

$$x+3y+2z$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}x&y&z\\3&-1&0\\2&0&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}x&y&z&x&y\\3&-1&0&3&-1\\2&0&-1&2&0\end{matrix}\right)$$

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

$$x\left(-1\right)\left(-1\right)=x$$

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

$$2\left(-1\right)z-3y=-3y-2z$$

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

$$x-\left(-3y-2z\right)$$

Subtract -2z-3y from x.

$$x+3y+2z$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}x&y&z\\3&-1&0\\2&0&-1\end{matrix}\right))$$

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.

$$xdet(\left(\begin{matrix}-1&0\\0&-1\end{matrix}\right))-ydet(\left(\begin{matrix}3&0\\2&-1\end{matrix}\right))+zdet(\left(\begin{matrix}3&-1\\2&0\end{matrix}\right))$$

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

$$x\left(-1\right)\left(-1\right)-y\times 3\left(-1\right)+z\left(-2\left(-1\right)\right)$$

Simplify.

$$x-y\left(-3\right)+z\times 2$$

Add the terms to obtain the final result.

$$x+3y+2z$$

Integrate w.r.t. x

$$2xz+3xy+\frac{x^{2}}{2}+С$$