Solve {ccc}{k}&{1}&{2}{1}&{k}&{-1}{2}&{-1}&{1}

$$\left| \begin{array} { c c c } { k } & { 1 } & { 2 } \\ { 1 } & { k } & { - 1 } \\ { 2 } & { - 1 } & { 1 } \end{array} \right|$$

Evaluate

$$k^{2}-5k-5$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$kk-2+2\left(-1\right)=k^{2}-4$$

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

$$2k\times 2-\left(-k\right)+1=5k+1$$

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

$$k^{2}-4-\left(5k+1\right)$$

Subtract 5k+1 from k^{2}-4.

$$k^{2}-5k-5$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$k\left(k-1\right)-3+2\left(-2k-1\right)$$

Add the terms to obtain the final result.

$$k^{2}-5k-5$$

Integrate w.r.t. k

$$\frac{k^{3}}{3}-\frac{5k^{2}}{2}-5k+С$$