Solve {ccc}{1}&{0}&{-2}{k}&{2}&{1}{4}&{k}&{5}

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

Evaluate

$$26-k-2k^{2}$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 5-2kk=10-2k^{2}$$

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

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

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

$$10-2k^{2}-\left(k-16\right)$$

Subtract -16+k from 10-2k^{2}.

$$26-k-2k^{2}$$

Steps Using Expansion by Minors

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

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

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.

$$det(\left(\begin{matrix}2&1\\k&5\end{matrix}\right))-2det(\left(\begin{matrix}k&2\\4&k\end{matrix}\right))$$

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

$$2\times 5-k-2\left(kk-4\times 2\right)$$

Simplify.

$$10-k-2\left(k^{2}-8\right)$$

Add the terms to obtain the final result.

$$26-k-2k^{2}$$

Integrate w.r.t. k

$$-\frac{2k^{3}}{3}-\frac{k^{2}}{2}+26k+С$$