Solve {ccc}{1}&{-1}&{a}{0}&{1}&{b}{1}&{0}&{c}

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

Evaluate

$$c-b-a$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$c-b=c-b$$

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

$$a=a$$

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

$$c-b-a$$

Steps Using Expansion by Minors

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

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

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.

$$det(\left(\begin{matrix}1&b\\0&c\end{matrix}\right))-\left(-det(\left(\begin{matrix}0&b\\1&c\end{matrix}\right))\right)+adet(\left(\begin{matrix}0&1\\1&0\end{matrix}\right))$$

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

$$c-\left(-\left(-b\right)\right)+a\left(-1\right)$$

Add the terms to obtain the final result.

$$c-b-a$$

Integrate w.r.t. c

$$-ac-bc+\frac{c^{2}}{2}+С$$