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

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

Evaluate

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

Steps Using Sarrus' Rule

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

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

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

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

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

$$kk+3\left(-1\right)k=k\left(k-3\right)$$

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

$$-2=-2$$

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

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

Subtract -2 from k\left(-3+k\right).

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

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

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

Integrate w.r.t. k

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