Solve {lll}{x}&{2}&{y}{8}&{9}&{3}{1}&{2}&{3}

$$\left| \begin{array} { l l l } { x } & { 2 } & { y } \\ { 8 } & { 9 } & { 3 } \\ { 1 } & { 2 } & { 3 } \end{array} \right|$$

Evaluate

$$7\left(3x+y-6\right)$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}x&2&y\\8&9&3\\1&2&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}x&2&y&x&2\\8&9&3&8&9\\1&2&3&1&2\end{matrix}\right)$$

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

$$x\times 9\times 3+2\times 3+y\times 8\times 2=27x+16y+6$$

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

$$9y+2\times 3x+3\times 8\times 2=6x+9y+48$$

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

$$27x+16y+6-\left(6x+9y+48\right)$$

Subtract 9y+6x+48 from 27x+6+16y.

$$21x+7y-42$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}x&2&y\\8&9&3\\1&2&3\end{matrix}\right))$$

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.

$$xdet(\left(\begin{matrix}9&3\\2&3\end{matrix}\right))-2det(\left(\begin{matrix}8&3\\1&3\end{matrix}\right))+ydet(\left(\begin{matrix}8&9\\1&2\end{matrix}\right))$$

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

$$x\left(9\times 3-2\times 3\right)-2\left(8\times 3-3\right)+y\left(8\times 2-9\right)$$

Simplify.

$$x\times 21-2\times 21+y\times 7$$

Add the terms to obtain the final result.

$$21x+7y-42$$

Integrate w.r.t. x

$$7xy+\frac{21x^{2}}{2}-42x+С$$