Solve {ccc}{m}&{n}&{p}{3}&{0}&{6}{1}&{3}&{2}

$$\left| \begin{array} { c c c } { m } & { n } & { p } \\ { 3 } & { 0 } & { 6 } \\ { 1 } & { 3 } & { 2 } \end{array} \right|$$

Evaluate

$$9\left(p-2m\right)$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}m&n&p\\3&0&6\\1&3&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}m&n&p&m&n\\3&0&6&3&0\\1&3&2&1&3\end{matrix}\right)$$

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

$$n\times 6+p\times 3\times 3=6n+9p$$

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

$$3\times 6m+2\times 3n=18m+6n$$

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

$$6n+9p-\left(18m+6n\right)$$

Subtract 18m+6n from 6n+9p.

$$9p-18m$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}m&n&p\\3&0&6\\1&3&2\end{matrix}\right))$$

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.

$$mdet(\left(\begin{matrix}0&6\\3&2\end{matrix}\right))-ndet(\left(\begin{matrix}3&6\\1&2\end{matrix}\right))+pdet(\left(\begin{matrix}3&0\\1&3\end{matrix}\right))$$

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

$$m\left(-3\times 6\right)-n\left(3\times 2-6\right)+p\times 3\times 3$$

Simplify.

$$m\left(-18\right)+p\times 9$$

Add the terms to obtain the final result.

$$9p-18m$$

Integrate w.r.t. p

$$-18mp+\frac{9p^{2}}{2}+С$$