Solve {lll}{3}&{4}&{1}{2}&{3}&{4}{4}&{1}&{2}

$$\left| \begin{array} { l l l } { 3 } & { 4 } & { 1 } \\ { 2 } & { 3 } & { 4 } \\ { 4 } & { 1 } & { 2 } \end{array} \right|$$

Evaluate

$$44$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\times 2+4\times 4\times 4+2=84$$

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

$$4\times 3+4\times 3+2\times 2\times 4=40$$

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

$$84-40$$

Subtract 40 from 84.

$$44$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 2-4\left(-12\right)-10$$

Add the terms to obtain the final result.

$$44$$

Factor

$$2^{2}\times 11$$