Solve {lll}{0}&{a}&{b}{a}&{0}&{c}{b}&{c}&{0}

$$\left| \begin{array} { l l l } { 0 } & { a } & { b } \\ { a } & { 0 } & { c } \\ { b } & { c } & { 0 } \end{array} \right|$$

Evaluate

$$2abc$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$acb+bac=2abc$$

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

$$\text{true}$$

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

$$2abc$$

Steps Using Expansion by Minors

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

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

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.

$$-adet(\left(\begin{matrix}a&c\\b&0\end{matrix}\right))+bdet(\left(\begin{matrix}a&0\\b&c\end{matrix}\right))$$

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

$$-a\left(-bc\right)+bac$$

Add the terms to obtain the final result.

$$2abc$$

Integrate w.r.t. a

$$С+bca^{2}$$