Solve {ccc}{m}&{n}&{p}{1}&{1}&{-1}{1}&{-1}&{1}

$$\left| \begin{array} { c c c } { m } & { n } & { p } \\ { 1 } & { 1 } & { - 1 } \\ { 1 } & { - 1 } & { 1 } \end{array} \right|$$

Evaluate

$$-2n-2p$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}m&n&p\\1&1&-1\\1&-1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}m&n&p&m&n\\1&1&-1&1&1\\1&-1&1&1&-1\end{matrix}\right)$$

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

$$m+n\left(-1\right)+p\left(-1\right)=m-n-p$$

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

$$p-\left(-m\right)+n=m+n+p$$

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

$$m-n-p-\left(m+n+p\right)$$

Subtract p+m+n from m-n-p.

$$-2n-2p$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}m&n&p\\1&1&-1\\1&-1&1\end{matrix}\right))$$

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.

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

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

$$m\left(1-\left(-\left(-1\right)\right)\right)-n\left(1-\left(-1\right)\right)+p\left(-1-1\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$-2n-2p$$

Integrate w.r.t. n

$$С-2np-n^{2}$$