Solve {ccc}{a}&{b}&{c}{5}&{-2}&{2}{3}&{0}&{2}

$$\left| \begin{array} { c c c } { a } & { b } & { c } \\ { 5 } & { - 2 } & { 2 } \\ { 3 } & { 0 } & { 2 } \end{array} \right|$$

Evaluate

$$6c-4b-4a$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}a&b&c\\5&-2&2\\3&0&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}a&b&c&a&b\\5&-2&2&5&-2\\3&0&2&3&0\end{matrix}\right)$$

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

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

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

$$3\left(-2\right)c+2\times 5b=10b-6c$$

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

$$6b-4a-\left(10b-6c\right)$$

Subtract -6c+10b from -4a+6b.

$$6c-4b-4a$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}a&b&c\\5&-2&2\\3&0&2\end{matrix}\right))$$

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.

$$adet(\left(\begin{matrix}-2&2\\0&2\end{matrix}\right))-bdet(\left(\begin{matrix}5&2\\3&2\end{matrix}\right))+cdet(\left(\begin{matrix}5&-2\\3&0\end{matrix}\right))$$

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

$$a\left(-2\right)\times 2-b\left(5\times 2-3\times 2\right)+c\left(-3\left(-2\right)\right)$$

Simplify.

$$a\left(-4\right)-b\times 4+c\times 6$$

Add the terms to obtain the final result.

$$6c-4b-4a$$

Integrate w.r.t. a

$$С-2a^{2}-4ab+6ac$$