Solve {ccc}{1!}&{2!}&{3!}{2!}&{3!}&{4!}{3!}&{4!}&{5!}=

$$\left| \begin{array} { c c c } { 1 ! } & { 2 ! } & { 3 ! } \\ { 2 ! } & { 3 ! } & { 4 ! } \\ { 3 ! } & { 4 ! } & { 5 ! } \end{array} \right| =$$

Evaluate

$$24$$

Steps Using Sarrus' Rule

The factorial of 1 is 1.

$$det(\left(\begin{matrix}1&2!&3!\\2!&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 2 is 2.

$$det(\left(\begin{matrix}1&2&3!\\2!&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 3 is 6.

$$det(\left(\begin{matrix}1&2&6\\2!&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 2 is 2.

$$det(\left(\begin{matrix}1&2&6\\2&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 3 is 6.

$$det(\left(\begin{matrix}1&2&6\\2&6&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 4 is 24.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\3!&4!&5!\end{matrix}\right))$$

The factorial of 3 is 6.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\6&4!&5!\end{matrix}\right))$$

The factorial of 4 is 24.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\6&24&5!\end{matrix}\right))$$

The factorial of 5 is 120.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\6&24&120\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&6&1&2\\2&6&24&2&6\\6&24&120&6&24\end{matrix}\right)$$

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

$$6\times 120+2\times 24\times 6+6\times 2\times 24=1296$$

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

$$6\times 6\times 6+24\times 24+120\times 2\times 2=1272$$

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

$$1296-1272$$

Subtract 1272 from 1296.

$$24$$

Steps Using Expansion by Minors

The factorial of 1 is 1.

$$det(\left(\begin{matrix}1&2!&3!\\2!&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 2 is 2.

$$det(\left(\begin{matrix}1&2&3!\\2!&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 3 is 6.

$$det(\left(\begin{matrix}1&2&6\\2!&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 2 is 2.

$$det(\left(\begin{matrix}1&2&6\\2&3!&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 3 is 6.

$$det(\left(\begin{matrix}1&2&6\\2&6&4!\\3!&4!&5!\end{matrix}\right))$$

The factorial of 4 is 24.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\3!&4!&5!\end{matrix}\right))$$

The factorial of 3 is 6.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\6&4!&5!\end{matrix}\right))$$

The factorial of 4 is 24.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\6&24&5!\end{matrix}\right))$$

The factorial of 5 is 120.

$$det(\left(\begin{matrix}1&2&6\\2&6&24\\6&24&120\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}6&24\\24&120\end{matrix}\right))-2det(\left(\begin{matrix}2&24\\6&120\end{matrix}\right))+6det(\left(\begin{matrix}2&6\\6&24\end{matrix}\right))$$

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

$$6\times 120-24\times 24-2\left(2\times 120-6\times 24\right)+6\left(2\times 24-6\times 6\right)$$

Simplify.

$$144-2\times 96+6\times 12$$

Add the terms to obtain the final result.

$$24$$

Factor

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