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

$$\left( \begin{array} { l l l } { 2 } & { 3 } & { 1 } \\ { 1 } & { 3 } & { 3 } \\ { 2 } & { 6 } & { 6 } \end{array} \right)$$

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

$$60-60$$

Subtract 60 from 60.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$0$$

Evaluate

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