Solve {lll}{6}&{a}&{b}{2}&{3}&{4}{3}&{4}&{9}

$$\left| \begin{array} { l l l } { 6 } & { a } & { b } \\ { 2 } & { 3 } & { 4 } \\ { 3 } & { 4 } & { 9 } \end{array} \right|$$

Evaluate

$$66-b-6a$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}6&a&b\\2&3&4\\3&4&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}6&a&b&6&a\\2&3&4&2&3\\3&4&9&3&4\end{matrix}\right)$$

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

$$6\times 3\times 9+a\times 4\times 3+b\times 2\times 4=12a+8b+162$$

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

$$3\times 3b+4\times 4\times 6+9\times 2a=18a+9b+96$$

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

$$12a+8b+162-\left(18a+9b+96\right)$$

Subtract 9b+96+18a from 162+12a+8b.

$$66-b-6a$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}6&a&b\\2&3&4\\3&4&9\end{matrix}\right))$$

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.

$$6det(\left(\begin{matrix}3&4\\4&9\end{matrix}\right))-adet(\left(\begin{matrix}2&4\\3&9\end{matrix}\right))+bdet(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))$$

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

$$6\left(3\times 9-4\times 4\right)-a\left(2\times 9-3\times 4\right)+b\left(2\times 4-3\times 3\right)$$

Simplify.

$$6\times 11-a\times 6+b\left(-1\right)$$

Add the terms to obtain the final result.

$$66-b-6a$$

Integrate w.r.t. a

$$С-3a^{2}-ab+66a$$