Solve {lll}{1}&{1}&{1}{3}&{3^2}&{3^3}{4}&{4^2}&{4^3}

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

Evaluate

$$72$$

Steps Using Sarrus' Rule

Calculate 3 to the power of 2 and get 9.

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

Calculate 3 to the power of 3 and get 27.

$$det(\left(\begin{matrix}1&1&1\\3&9&27\\4&4^{2}&4^{3}\end{matrix}\right))$$

Calculate 4 to the power of 2 and get 16.

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

Calculate 4 to the power of 3 and get 64.

$$det(\left(\begin{matrix}1&1&1\\3&9&27\\4&16&64\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\3&9&27&3&9\\4&16&64&4&16\end{matrix}\right)$$

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

$$9\times 64+27\times 4+3\times 16=732$$

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

$$4\times 9+16\times 27+64\times 3=660$$

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

$$732-660$$

Subtract 660 from 732.

$$72$$

Steps Using Expansion by Minors

Calculate 3 to the power of 2 and get 9.

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

Calculate 3 to the power of 3 and get 27.

$$det(\left(\begin{matrix}1&1&1\\3&9&27\\4&4^{2}&4^{3}\end{matrix}\right))$$

Calculate 4 to the power of 2 and get 16.

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

Calculate 4 to the power of 3 and get 64.

$$det(\left(\begin{matrix}1&1&1\\3&9&27\\4&16&64\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}9&27\\16&64\end{matrix}\right))-det(\left(\begin{matrix}3&27\\4&64\end{matrix}\right))+det(\left(\begin{matrix}3&9\\4&16\end{matrix}\right))$$

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

$$9\times 64-16\times 27-\left(3\times 64-4\times 27\right)+3\times 16-4\times 9$$

Simplify.

$$144-84+12$$

Add the terms to obtain the final result.

$$72$$

Factor

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