Solve {lll}{lambda}&{j}&{k}{1}&{0}&{2}{0}&{1}&{1}

$$\left| \begin{array} { l l l } { \lambda } & { j } & { k } \\ { 1 } & { 0 } & { 2 } \\ { 0 } & { 1 } & { 1 } \end{array} \right|$$

Evaluate

$$-j+k-2\lambda$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$k=k$$

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

$$2\lambda +j=j+2\lambda$$

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

$$k-\left(j+2\lambda \right)$$

Subtract 2\lambda +j from k.

$$-j+k-2\lambda$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$\lambda \left(-2\right)-j+k$$

Add the terms to obtain the final result.

$$-j+k-2\lambda$$

Integrate w.r.t. k

$$-jk-2k\lambda +\frac{k^{2}}{2}+С$$