Solve {ccc}{a}&{b}&{c}{1}&{0}&{-3}{2}&{-4}&{-2}

$$\left| \begin{array} { c c c } { a } & { b } & { c } \\ { 1 } & { 0 } & { - 3 } \\ { 2 } & { - 4 } & { - 2 } \end{array} \right|$$

Evaluate

$$-12a-4b-4c$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}a&b&c\\1&0&-3\\2&-4&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}a&b&c&a&b\\1&0&-3&1&0\\2&-4&-2&2&-4\end{matrix}\right)$$

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

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

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

$$-4\left(-3\right)a-2b=12a-2b$$

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

$$-6b-4c-\left(12a-2b\right)$$

Subtract 12a-2b from -6b-4c.

$$-12a-4b-4c$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}a&b&c\\1&0&-3\\2&-4&-2\end{matrix}\right))$$

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.

$$adet(\left(\begin{matrix}0&-3\\-4&-2\end{matrix}\right))-bdet(\left(\begin{matrix}1&-3\\2&-2\end{matrix}\right))+cdet(\left(\begin{matrix}1&0\\2&-4\end{matrix}\right))$$

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

$$a\left(-\left(-4\left(-3\right)\right)\right)-b\left(-2-2\left(-3\right)\right)+c\left(-4\right)$$

Simplify.

$$a\left(-12\right)-b\times 4+c\left(-4\right)$$

Add the terms to obtain the final result.

$$-12a-4b-4c$$

Integrate w.r.t. b

$$С-12ab-2b^{2}-4bc$$