Solve {rrr}{k}&{-1}&{1}{2}&{2}&{-6}{1}&{-3}&{4}=

$$\left| \begin{array} { r r r } { k } & { - 1 } & { 1 } \\ { 2 } & { 2 } & { - 6 } \\ { 1 } & { - 3 } & { 4 } \end{array} \right| =$$

Evaluate

$$6-10k$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}k&-1&1\\2&2&-6\\1&-3&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}k&-1&1&k&-1\\2&2&-6&2&2\\1&-3&4&1&-3\end{matrix}\right)$$

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

$$k\times 2\times 4-\left(-6\right)+2\left(-3\right)=8k$$

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

$$2-3\left(-6\right)k+4\times 2\left(-1\right)=18k-6$$

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

$$8k-\left(18k-6\right)$$

Subtract -6+18k from 8k.

$$6-10k$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}k&-1&1\\2&2&-6\\1&-3&4\end{matrix}\right))$$

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.

$$kdet(\left(\begin{matrix}2&-6\\-3&4\end{matrix}\right))-\left(-det(\left(\begin{matrix}2&-6\\1&4\end{matrix}\right))\right)+det(\left(\begin{matrix}2&2\\1&-3\end{matrix}\right))$$

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

$$k\left(2\times 4-\left(-3\left(-6\right)\right)\right)-\left(-\left(2\times 4-\left(-6\right)\right)\right)+2\left(-3\right)-2$$

Simplify.

$$k\left(-10\right)-\left(-14\right)-8$$

Add the terms to obtain the final result.

$$6-10k$$

Integrate w.r.t. k

$$С-5k^{2}+6k$$