Question

The value of the determinant
$\Delta =\left|\begin{array}{c}1!\\ 2!\\ 3!\end{array}\begin{array}{c}2!\\ 3!\\ 4!\end{array}\begin{array}{c}3!\\ 4!\\ 5!\end{array}\right|$ is

• A. $2!$
• B. $3!$
• C. $4!$
• D. $5!$

Answer 1

The correct option is C $4!$

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

The above matrix can be rewritten as:

$\Delta =\left|\begin{array}{ccc}1\times 1& 2\times 1& 3\times 2\times 1\\ 2\times 1& 3\times 2\times 1& 4\times 3\times 2\times 1\\ 3\times 2\times 1& 4\times 3\times 2\times 1& 5\times 4\times 3\times 2\times 1\end{array}\right|$

$\Delta =\left|\begin{array}{ccc}1& 2& 6\\ 2& 6& 24\\ 6& 24& 120\end{array}\right|$

Expand the above determinant along $R_1$ as shown below:

$\Delta =1[\left(6\right)\left(120\right)-\left(24\right)\left(24\right)]-2[\left(2\right)\left(120\right)-\left(6\right)\left(24\right)]+6[\left(2\right)\left(24\right)-\left(6\right)\left(6\right)]$

$\Delta =1[720-576]-2[240-144]+6[48-36]$

$\Delta =1\left(144\right)-2\left(96\right)+6\left(12\right)$

$\Delta =144-192+72$

$\Delta =24$

$\Delta =4!$

Hence option $C$ is correct.