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

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

Evaluate

$$x-2$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2\times 2\times 2+3x+2y=3x+2y+8$$

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

$$4x+2y+6-\left(3x+2y+8\right)$$

Subtract 8+3x+2y from 6+4x+2y.

$$x-2$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$x-2$$

Integrate w.r.t. x

$$\frac{x^{2}}{2}-2x+С$$