Solve {ccc}{A}&{B}&{Z}{7}&{-6}&{1}{1}&{-2}&{1}

$$\left| \begin{array} { c c c } { A } & { B } & { Z } \\ { 7 } & { - 6 } & { 1 } \\ { 1 } & { - 2 } & { 1 } \end{array} \right|$$

Evaluate

$$-4A-6B-8Z$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}A&B&Z\\7&-6&1\\1&-2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}A&B&Z&A&B\\7&-6&1&7&-6\\1&-2&1&1&-2\end{matrix}\right)$$

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

$$A\left(-6\right)+B+Z\times 7\left(-2\right)=-6A+B-14Z$$

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

$$-6Z-2A+7B=-2A+7B-6Z$$

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

$$-6A+B-14Z-\left(-2A+7B-6Z\right)$$

Subtract -6Z-2A+7B from -6A+B-14Z.

$$-4A-6B-8Z$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}A&B&Z\\7&-6&1\\1&-2&1\end{matrix}\right))$$

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.

$$Adet(\left(\begin{matrix}-6&1\\-2&1\end{matrix}\right))-Bdet(\left(\begin{matrix}7&1\\1&1\end{matrix}\right))+Zdet(\left(\begin{matrix}7&-6\\1&-2\end{matrix}\right))$$

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

$$A\left(-6-\left(-2\right)\right)-B\left(7-1\right)+Z\left(7\left(-2\right)-\left(-6\right)\right)$$

Simplify.

$$A\left(-4\right)-B\times 6+Z\left(-8\right)$$

Add the terms to obtain the final result.

$$-4A-6B-8Z$$

Integrate w.r.t. A

$$С-2A^{2}-6AB-8AZ$$