Solve {rrr}{2}&{-1}&{4}{4}&{1}&{16}{8}&{-1}&{48}

$$\left| \begin{array} { r r r } { 2 } & { - 1 } & { 4 } \\ { 4 } & { 1 } & { 16 } \\ { 8 } & { - 1 } & { 48 } \end{array} \right|$$

Evaluate

$$144$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&4\\4&1&16\\8&-1&48\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&4&2&-1\\4&1&16&4&1\\8&-1&48&8&-1\end{matrix}\right)$$

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

$$2\times 48-16\times 8+4\times 4\left(-1\right)=-48$$

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

$$8\times 4-16\times 2+48\times 4\left(-1\right)=-192$$

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

$$-48-\left(-192\right)$$

Subtract -192 from -48.

$$144$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&4\\4&1&16\\8&-1&48\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&16\\-1&48\end{matrix}\right))-\left(-det(\left(\begin{matrix}4&16\\8&48\end{matrix}\right))\right)+4det(\left(\begin{matrix}4&1\\8&-1\end{matrix}\right))$$

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

$$2\left(48-\left(-16\right)\right)-\left(-\left(4\times 48-8\times 16\right)\right)+4\left(4\left(-1\right)-8\right)$$

Simplify.

$$2\times 64-\left(-64\right)+4\left(-12\right)$$

Add the terms to obtain the final result.

$$144$$

Factor

$$2^{4}\times 3^{2}$$