Solve {ccc}{1}&{1}&{-2}{7}&{0}&{3}{7}&{0}&{-6}

$$\left| \begin{array} { c c c } { 1 } & { 1 } & { - 2 } \\ { 7 } & { 0 } & { 3 } \\ { 7 } & { 0 } & { - 6 } \end{array} \right|$$

Evaluate

$$63$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&-2\\7&0&3\\7&0&-6\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&-2&1&1\\7&0&3&7&0\\7&0&-6&7&0\end{matrix}\right)$$

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

$$3\times 7=21$$

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

$$-6\times 7=-42$$

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

$$21-\left(-42\right)$$

Subtract -42 from 21.

$$63$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&-2\\7&0&3\\7&0&-6\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}0&3\\0&-6\end{matrix}\right))-det(\left(\begin{matrix}7&3\\7&-6\end{matrix}\right))-2det(\left(\begin{matrix}7&0\\7&0\end{matrix}\right))$$

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

$$-\left(7\left(-6\right)-7\times 3\right)$$

Simplify.

$$-\left(-63\right)$$

Add the terms to obtain the final result.

$$63$$

Factor

$$3^{2}\times 7$$