Solve {lll}{1}&{2}&{1}{2}&{4}&{6}{4}&{1}&{1}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 1 } \\ { 2 } & { 4 } & { 6 } \\ { 4 } & { 1 } & { 1 } \end{array} \right|$$

Evaluate

$$28$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&1\\2&4&6\\4&1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&1&1&2\\2&4&6&2&4\\4&1&1&4&1\end{matrix}\right)$$

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

$$4+2\times 6\times 4+2=54$$

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

$$4\times 4+6+2\times 2=26$$

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

$$54-26$$

Subtract 26 from 54.

$$28$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&1\\2&4&6\\4&1&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&6\\1&1\end{matrix}\right))-2det(\left(\begin{matrix}2&6\\4&1\end{matrix}\right))+det(\left(\begin{matrix}2&4\\4&1\end{matrix}\right))$$

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

$$4-6-2\left(2-4\times 6\right)+2-4\times 4$$

Simplify.

$$-2-2\left(-22\right)-14$$

Add the terms to obtain the final result.

$$28$$

Factor

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