Solve {rrr}{2}&{-2}&{1}{1}&{2}&{x}{1}&{1}&{1}

$$\left| \begin{array} { r r r } { 2 } & { - 2 } & { 1 } \\ { 1 } & { 2 } & { x } \\ { 1 } & { 1 } & { 1 } \end{array} \right|$$

Evaluate

$$5-4x$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2-2x+1=5-2x$$

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

$$2+x\times 2-2=2x$$

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

$$5-2x-2x$$

Subtract 2x from 5-2x.

$$5-4x$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$5-4x$$

Integrate w.r.t. x

$$С+5x-2x^{2}$$