Solve {ccc}{2}&{2}&{k}{0}&{1}&{1}{-3}&{1}&{0}

$$\left| \begin{array} { c c c } { 2 } & { 2 } & { k } \\ { 0 } & { 1 } & { 1 } \\ { - 3 } & { 1 } & { 0 } \end{array} \right|$$

Evaluate

$$3k-8$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&2&k\\0&1&1\\-3&1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&2&k&2&2\\0&1&1&0&1\\-3&1&0&-3&1\end{matrix}\right)$$

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

$$2\left(-3\right)=-6$$

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

$$-3k+2=2-3k$$

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

$$-6-\left(2-3k\right)$$

Subtract -3k+2 from -6.

$$3k-8$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&2&k\\0&1&1\\-3&1&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&1\\1&0\end{matrix}\right))-2det(\left(\begin{matrix}0&1\\-3&0\end{matrix}\right))+kdet(\left(\begin{matrix}0&1\\-3&1\end{matrix}\right))$$

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

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

Simplify.

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

Add the terms to obtain the final result.

$$3k-8$$

Integrate w.r.t. k

$$\frac{3k^{2}}{2}-8k+С$$