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

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

Evaluate

$$63$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 3+3\times 2\times 3+4\times 5\times 6=144$$

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

$$3\times 4+6\times 2\times 2+3\times 5\times 3=81$$

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

$$144-81$$

Subtract 81 from 144.

$$63$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-9\right)-3\times 9+4\times 27$$

Add the terms to obtain the final result.

$$63$$

Factor

$$3^{2}\times 7$$