Solve {ccc}{16}&{10}&{4}{10}&{20}&{8}{4}&{8}&{12}

$$\left| \begin{array} { c c c } { 16 } & { 10 } & { 4 } \\ { 10 } & { 20 } & { 8 } \\ { 4 } & { 8 } & { 12 } \end{array} \right|$$

Evaluate

$$1936$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}16&10&4\\10&20&8\\4&8&12\end{matrix}\right))$$

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

$$\left(\begin{matrix}16&10&4&16&10\\10&20&8&10&20\\4&8&12&4&8\end{matrix}\right)$$

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

$$16\times 20\times 12+10\times 8\times 4+4\times 10\times 8=4480$$

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

$$4\times 20\times 4+8\times 8\times 16+12\times 10\times 10=2544$$

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

$$4480-2544$$

Subtract 2544 from 4480.

$$1936$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}16&10&4\\10&20&8\\4&8&12\end{matrix}\right))$$

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.

$$16det(\left(\begin{matrix}20&8\\8&12\end{matrix}\right))-10det(\left(\begin{matrix}10&8\\4&12\end{matrix}\right))+4det(\left(\begin{matrix}10&20\\4&8\end{matrix}\right))$$

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

$$16\left(20\times 12-8\times 8\right)-10\left(10\times 12-4\times 8\right)+4\left(10\times 8-4\times 20\right)$$

Simplify.

$$16\times 176-10\times 88$$

Add the terms to obtain the final result.

$$1936$$

Factor

$$2^{4}\times 11^{2}$$