Solve {ccc}{2}&{4}&{9}{3}&{9}&{28}{4}&{16}&{65}

$$\left| \begin{array} { c c c } { 2 } & { 4 } & { 9 } \\ { 3 } & { 9 } & { 28 } \\ { 4 } & { 16 } & { 65 } \end{array} \right|$$

Evaluate

$$50$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&4&9\\3&9&28\\4&16&65\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&4&9&2&4\\3&9&28&3&9\\4&16&65&4&16\end{matrix}\right)$$

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

$$2\times 9\times 65+4\times 28\times 4+9\times 3\times 16=2050$$

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

$$4\times 9\times 9+16\times 28\times 2+65\times 3\times 4=2000$$

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

$$2050-2000$$

Subtract 2000 from 2050.

$$50$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&4&9\\3&9&28\\4&16&65\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}9&28\\16&65\end{matrix}\right))-4det(\left(\begin{matrix}3&28\\4&65\end{matrix}\right))+9det(\left(\begin{matrix}3&9\\4&16\end{matrix}\right))$$

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

$$2\left(9\times 65-16\times 28\right)-4\left(3\times 65-4\times 28\right)+9\left(3\times 16-4\times 9\right)$$

Simplify.

$$2\times 137-4\times 83+9\times 12$$

Add the terms to obtain the final result.

$$50$$

Factor

$$2\times 5^{2}$$