Solve {ccc}{1}&{2}&{3}{0}&{5}&{-4}{2}&{8}&{x}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { 3 } \\ { 0 } & { 5 } & { - 4 } \\ { 2 } & { 8 } & { x } \end{array} \right|$$

Evaluate

$$5x-14$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\0&5&-4\\2&8&x\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\0&5&-4&0&5\\2&8&x&2&8\end{matrix}\right)$$

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

$$5x+2\left(-4\right)\times 2=5x-16$$

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

$$2\times 5\times 3+8\left(-4\right)=-2$$

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

$$5x-16-\left(-2\right)$$

Subtract -2 from 5x-16.

$$5x-14$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\0&5&-4\\2&8&x\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}5&-4\\8&x\end{matrix}\right))-2det(\left(\begin{matrix}0&-4\\2&x\end{matrix}\right))+3det(\left(\begin{matrix}0&5\\2&8\end{matrix}\right))$$

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

$$5x-8\left(-4\right)-2\left(-2\left(-4\right)\right)+3\left(-2\times 5\right)$$

Simplify.

$$5x+32-2\times 8+3\left(-10\right)$$

Add the terms to obtain the final result.

$$5x-14$$

Integrate w.r.t. x

$$\frac{5x^{2}}{2}-14x+С$$