Solve {ccc}{-2}&{-2}&{2}{-7}&{-10}&{-12}{0}&{-3}&{-9}

$$\left| \begin{array} { c c c } { - 2 } & { - 2 } & { 2 } \\ { - 7 } & { - 10 } & { - 12 } \\ { 0 } & { - 3 } & { - 9 } \end{array} \right|$$

Evaluate

$$60$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&-2&2\\-7&-10&-12\\0&-3&-9\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&-2&2&-2&-2\\-7&-10&-12&-7&-10\\0&-3&-9&0&-3\end{matrix}\right)$$

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

$$-2\left(-10\right)\left(-9\right)+2\left(-7\right)\left(-3\right)=-138$$

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

$$-3\left(-12\right)\left(-2\right)-9\left(-7\right)\left(-2\right)=-198$$

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

$$-138-\left(-198\right)$$

Subtract -198 from -138.

$$60$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&-2&2\\-7&-10&-12\\0&-3&-9\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}-10&-12\\-3&-9\end{matrix}\right))-\left(-2det(\left(\begin{matrix}-7&-12\\0&-9\end{matrix}\right))\right)+2det(\left(\begin{matrix}-7&-10\\0&-3\end{matrix}\right))$$

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

$$-2\left(-10\left(-9\right)-\left(-3\left(-12\right)\right)\right)-\left(-2\left(-7\right)\left(-9\right)\right)+2\left(-7\right)\left(-3\right)$$

Simplify.

$$-2\times 54-\left(-2\times 63\right)+2\times 21$$

Add the terms to obtain the final result.

$$60$$

Factor

$$2^{2}\times 3\times 5$$