Solve {lll}{4}&{2}&{1}{1}&{3}&{5}{6}&{2}&{1}

$$\left| \begin{array} { l l l } { 4 } & { 2 } & { 1 } \\ { 1 } & { 3 } & { 5 } \\ { 6 } & { 2 } & { 1 } \end{array} \right|$$

Evaluate

$$14$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&2&1\\1&3&5\\6&2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&2&1&4&2\\1&3&5&1&3\\6&2&1&6&2\end{matrix}\right)$$

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

$$4\times 3+2\times 5\times 6+2=74$$

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

$$6\times 3+2\times 5\times 4+2=60$$

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

$$74-60$$

Subtract 60 from 74.

$$14$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&2&1\\1&3&5\\6&2&1\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}3&5\\2&1\end{matrix}\right))-2det(\left(\begin{matrix}1&5\\6&1\end{matrix}\right))+det(\left(\begin{matrix}1&3\\6&2\end{matrix}\right))$$

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

$$4\left(3-2\times 5\right)-2\left(1-6\times 5\right)+2-6\times 3$$

Simplify.

$$4\left(-7\right)-2\left(-29\right)-16$$

Add the terms to obtain the final result.

$$14$$

Factor

$$2\times 7$$