Solve operatorname{det}{lll}{x}&{x}&{y}{2}&{1}&{3}{1}&{2}&{3}

$$\operatorname { det } \left( \begin{array} { l l l } { x } & { x } & { y } \\ { 2 } & { 1 } & { 3 } \\ { 1 } & { 2 } & { 3 } \end{array} \right)$$

Evaluate

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

Steps Using Sarrus' Rule

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

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

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

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

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

$$x\times 3+x\times 3+y\times 2\times 2=6x+4y$$

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

$$y+2\times 3x+3\times 2x=12x+y$$

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

$$6x+4y-\left(12x+y\right)$$

Subtract y+12x from 6x+4y.

$$3y-6x$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$3y-6x$$

Integrate w.r.t. x

$$С+3xy-3x^{2}$$