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

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

Evaluate

$$-a^{3}-b^{3}$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$\text{true}$$

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

$$aaa+bbb=a^{3}+b^{3}$$

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

$$-\left(a^{3}+b^{3}\right)$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$a\left(-aa\right)-bbb$$

Simplify.

$$a\left(-a^{2}\right)-bb^{2}$$

Add the terms to obtain the final result.

$$-a^{3}-b^{3}$$

Integrate w.r.t. a

$$-ab^{3}-\frac{a^{4}}{4}+С$$