Solve {lll}{1}&{1}&{1}{4}&{2}&{3}{16}&{4}&{9}

$$\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 4 } & { 2 } & { 3 } \\ { 16 } & { 4 } & { 9 } \end{array} \right|$$

Evaluate

$$2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\4&2&3\\16&4&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\4&2&3&4&2\\16&4&9&16&4\end{matrix}\right)$$

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

$$2\times 9+3\times 16+4\times 4=82$$

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

$$16\times 2+4\times 3+9\times 4=80$$

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

$$82-80$$

Subtract 80 from 82.

$$2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\4&2&3\\16&4&9\end{matrix}\right))$$

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.

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

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

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

Simplify.

$$6-\left(-12\right)-16$$

Add the terms to obtain the final result.

$$2$$

Factor

$$2$$