Solve {ccc}{b}&{a}&{0}{e}&{c}&{0}{0}&{-2}&{-1}

$$\left| \begin{array} { c c c } { b } & { a } & { 0 } \\ { e } & { c } & { 0 } \\ { 0 } & { - 2 } & { - 1 } \end{array} \right|$$

Evaluate

$$ea-bc$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}b&a&0\\e&c&0\\0&-2&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}b&a&0&b&a\\e&c&0&e&c\\0&-2&-1&0&-2\end{matrix}\right)$$

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

$$bc\left(-1\right)=-bc$$

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

$$-ea=-ea$$

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

$$-bc-\left(-ea\right)$$

Subtract -ea from -bc.

$$ea-bc$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}b&a&0\\e&c&0\\0&-2&-1\end{matrix}\right))$$

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.

$$bdet(\left(\begin{matrix}c&0\\-2&-1\end{matrix}\right))-adet(\left(\begin{matrix}e&0\\0&-1\end{matrix}\right))$$

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

$$bc\left(-1\right)-ae\left(-1\right)$$

Simplify.

$$b\left(-c\right)-a\left(-e\right)$$

Add the terms to obtain the final result.

$$ea-bc$$

Integrate w.r.t. b

$$-\frac{cb^{2}}{2}+eab+С$$