Solve {rrr}{-1}&{3}&{4}{1}&{9}&{16}{-1}&{27}&{48}

$$\left| \begin{array} { r r r } { - 1 } & { 3 } & { 4 } \\ { 1 } & { 9 } & { 16 } \\ { - 1 } & { 27 } & { 48 } \end{array} \right|$$

Evaluate

$$-48$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-1&3&4\\1&9&16\\-1&27&48\end{matrix}\right))$$

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

$$\left(\begin{matrix}-1&3&4&-1&3\\1&9&16&1&9\\-1&27&48&-1&27\end{matrix}\right)$$

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

$$-9\times 48+3\times 16\left(-1\right)+4\times 27=-372$$

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

$$-9\times 4+27\times 16\left(-1\right)+48\times 3=-324$$

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

$$-372-\left(-324\right)$$

Subtract -324 from -372.

$$-48$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-1&3&4\\1&9&16\\-1&27&48\end{matrix}\right))$$

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.

$$-det(\left(\begin{matrix}9&16\\27&48\end{matrix}\right))-3det(\left(\begin{matrix}1&16\\-1&48\end{matrix}\right))+4det(\left(\begin{matrix}1&9\\-1&27\end{matrix}\right))$$

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

$$-\left(9\times 48-27\times 16\right)-3\left(48-\left(-16\right)\right)+4\left(27-\left(-9\right)\right)$$

Simplify.

$$-3\times 64+4\times 36$$

Add the terms to obtain the final result.

$$-48$$

Factor

$$-48$$