Solve {lll}{2}&{4}&{8}{3}&{9}&{2}{4}&{16}&{64}

$$\left| \begin{array} { l l l } { 2 } & { 4 } & { 8 } \\ { 3 } & { 9 } & { 2 } \\ { 4 } & { 16 } & { 64 } \end{array} \right|$$

Evaluate

$$448$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&4&8\\3&9&2\\4&16&64\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&4&8&2&4\\3&9&2&3&9\\4&16&64&4&16\end{matrix}\right)$$

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

$$2\times 9\times 64+4\times 2\times 4+8\times 3\times 16=1568$$

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

$$4\times 9\times 8+16\times 2\times 2+64\times 3\times 4=1120$$

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

$$1568-1120$$

Subtract 1120 from 1568.

$$448$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&4&8\\3&9&2\\4&16&64\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}9&2\\16&64\end{matrix}\right))-4det(\left(\begin{matrix}3&2\\4&64\end{matrix}\right))+8det(\left(\begin{matrix}3&9\\4&16\end{matrix}\right))$$

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

$$2\left(9\times 64-16\times 2\right)-4\left(3\times 64-4\times 2\right)+8\left(3\times 16-4\times 9\right)$$

Simplify.

$$2\times 544-4\times 184+8\times 12$$

Add the terms to obtain the final result.

$$448$$

Factor

$$2^{6}\times 7$$