Solve {lll}{x}&{y}&{z}{1}&{3}&{2}{1}&{4}&{5}

$$\left| \begin{array} { l l l } { x } & { y } & { z } \\ { 1 } & { 3 } & { 2 } \\ { 1 } & { 4 } & { 5 } \end{array} \right|$$

Evaluate

$$7x-3y+z$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}x&y&z\\1&3&2\\1&4&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}x&y&z&x&y\\1&3&2&1&3\\1&4&5&1&4\end{matrix}\right)$$

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

$$x\times 3\times 5+y\times 2+z\times 4=15x+2y+4z$$

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

$$3z+4\times 2x+5y=8x+5y+3z$$

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

$$15x+2y+4z-\left(8x+5y+3z\right)$$

Subtract 3z+8x+5y from 15x+2y+4z.

$$7x-3y+z$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}x&y&z\\1&3&2\\1&4&5\end{matrix}\right))$$

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.

$$xdet(\left(\begin{matrix}3&2\\4&5\end{matrix}\right))-ydet(\left(\begin{matrix}1&2\\1&5\end{matrix}\right))+zdet(\left(\begin{matrix}1&3\\1&4\end{matrix}\right))$$

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

$$x\left(3\times 5-4\times 2\right)-y\left(5-2\right)+z\left(4-3\right)$$

Simplify.

$$x\times 7-y\times 3+z$$

Add the terms to obtain the final result.

$$7x-3y+z$$

Integrate w.r.t. x

$$xz-3xy+\frac{7x^{2}}{2}+С$$