Solve {ccc}{2}&{3}&{1}{4}&{100}&{2}{2}&{-2}&{0}

$$\left| \begin{array} { c c c } { 2 } & { 3 } & { 1 } \\ { 4 } & { 100 } & { 2 } \\ { 2 } & { - 2 } & { 0 } \end{array} \right|$$

Evaluate

$$-188$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&1\\4&100&2\\2&-2&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&1&2&3\\4&100&2&4&100\\2&-2&0&2&-2\end{matrix}\right)$$

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

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

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

$$2\times 100-2\times 2\times 2=192$$

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

$$4-192$$

Subtract 192 from 4.

$$-188$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&1\\4&100&2\\2&-2&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}100&2\\-2&0\end{matrix}\right))-3det(\left(\begin{matrix}4&2\\2&0\end{matrix}\right))+det(\left(\begin{matrix}4&100\\2&-2\end{matrix}\right))$$

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

$$2\left(-\left(-2\times 2\right)\right)-3\left(-2\times 2\right)+4\left(-2\right)-2\times 100$$

Simplify.

$$2\times 4-3\left(-4\right)-208$$

Add the terms to obtain the final result.

$$-188$$

Factor

$$-188$$