Solve {ccc}{-2}&{1}&{10}{2}&{2}&{1}{3}&{4}&{4}

$$\left| \begin{array} { c c c } { - 2 } & { 1 } & { 10 } \\ { 2 } & { 2 } & { 1 } \\ { 3 } & { 4 } & { 4 } \end{array} \right|$$

Evaluate

$$7$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&1&10\\2&2&1\\3&4&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&1&10&-2&1\\2&2&1&2&2\\3&4&4&3&4\end{matrix}\right)$$

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

$$-2\times 2\times 4+3+10\times 2\times 4=67$$

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

$$3\times 2\times 10+4\left(-2\right)+4\times 2=60$$

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

$$67-60$$

Subtract 60 from 67.

$$7$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&1&10\\2&2&1\\3&4&4\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}2&1\\4&4\end{matrix}\right))-det(\left(\begin{matrix}2&1\\3&4\end{matrix}\right))+10det(\left(\begin{matrix}2&2\\3&4\end{matrix}\right))$$

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

$$-2\left(2\times 4-4\right)-\left(2\times 4-3\right)+10\left(2\times 4-3\times 2\right)$$

Simplify.

$$-2\times 4-5+10\times 2$$

Add the terms to obtain the final result.

$$7$$

Factor

$$7$$