Solve {ccc}{1}&{j}&{k}{-18}&{0}&{0}{1}&{5}&{-5}

$$\left| \begin{array} { c c c } { 1 } & { j } & { k } \\ { - 18 } & { 0 } & { 0 } \\ { 1 } & { 5 } & { - 5 } \end{array} \right|$$

Evaluate

$$-90j-90k$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&j&k\\-18&0&0\\1&5&-5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&j&k&1&j\\-18&0&0&-18&0\\1&5&-5&1&5\end{matrix}\right)$$

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

$$k\left(-18\right)\times 5=-90k$$

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

$$-5\left(-18\right)j=90j$$

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

$$-90k-90j$$

Subtract 90j from -90k.

$$-90j-90k$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&j&k\\-18&0&0\\1&5&-5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}0&0\\5&-5\end{matrix}\right))-jdet(\left(\begin{matrix}-18&0\\1&-5\end{matrix}\right))+kdet(\left(\begin{matrix}-18&0\\1&5\end{matrix}\right))$$

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

$$-j\left(-18\right)\left(-5\right)+k\left(-18\right)\times 5$$

Simplify.

$$-j\times 90+k\left(-90\right)$$

Add the terms to obtain the final result.

$$-90j-90k$$

Integrate w.r.t. k

$$С-90jk-45k^{2}$$