Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} 1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}\right|=?$ A. 2 B. 6 C. 24 D. 120

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Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} 1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}\right|=?$
A. 2
B. 6
C. 24
D. 120

CBSE | Class 12 | Determinants | Maths | RS Aggarwal

Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} 1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}\right|=?$
A. 2
B. 6
C. 24
D. 120

Solution:

Option(24)
To find: Value of $\left|\begin{array}{lll}1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 !\end{array}\right|$
We have, $\mid\left[\begin{array}{lll}1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 !\end{array} \mid\right.$
$\Rightarrow\left|\begin{array}{ccc} 1 & 2 & 6 \\ 2 & 6 & 24 \\ 6 & 24 & 120 \end{array}\right|$
Taking 2 common from $R_{2}$
$\Rightarrow 2\left|\begin{array}{ccc} 1 & 2 & 6 \\ 1 & 3 & 12 \\ 6 & 24 & 120 \end{array}\right|$
Taking 6 common from $R_{3}$
$\Rightarrow 2 \times 6\left|\begin{array}{ccc} 1 & 2 & 6 \\ 1 & 3 & 12 \\ 1 & 4 & 20 \end{array}\right|$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$
$\Rightarrow 12\left|\begin{array}{ccc} 1 & 2 & 6 \\ 0 & 1 & 6 \\ 1 & 4 & 20 \end{array}\right|$
Applying $R_{3} \rightarrow R_{3}-R_{1}$
$\Rightarrow 12\left|\begin{array}{ccc} 1 & 2 & 6 \\ 0 & 1 & 6 \\ 0 & 2 & 14 \end{array}\right|$
Expanding column 1
$\begin{array}{l} \Rightarrow 12[1\{(1)(14)-(6)(2)\}] \\ \Rightarrow 12[1\{(14)-(12)\}] \\ \Rightarrow 12[2] \end{array}$
$\Rightarrow 24$