Solve {lll}{a}&{1}&{1}{1}&{b}&{1}{1}&{2}&{b}

$$\left| \begin{array} { l l l } { a } & { 1 } & { 1 } \\ { 1 } & { b } & { 1 } \\ { 1 } & { 2 } & { b } \end{array} \right|$$

Evaluate

$$ab^{2}-2a-2b+3$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$abb+1+2=ab^{2}+3$$

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

$$b+2a+b=2a+2b$$

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

$$ab^{2}+3-\left(2a+2b\right)$$

Subtract 2b+2a from ab^{2}+3.

$$ab^{2}-2a-2b+3$$

Steps Using Expansion by Minors

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

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

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.

$$adet(\left(\begin{matrix}b&1\\2&b\end{matrix}\right))-det(\left(\begin{matrix}1&1\\1&b\end{matrix}\right))+det(\left(\begin{matrix}1&b\\1&2\end{matrix}\right))$$

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

$$a\left(bb-2\right)-\left(b-1\right)+2-b$$

Simplify.

$$a\left(b^{2}-2\right)-\left(b-1\right)+2-b$$

Add the terms to obtain the final result.

$$ab^{2}-2a-2b+3$$

Integrate w.r.t. b

$$\frac{ab^{3}}{3}-2ab-b^{2}+3b+С$$