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UPSC NDA 2 2019 Mathematics Solved Paper

Question : 11 of 120

Marks: +1, -0

\begin{vmatrix} 1! & 2! & 3! \\ 2! & 3! & 4! \\ 3! & 4! & 5! \end{vmatrix} ?

Solution:

Concept:

The determinant of a

3 \times 3

matrix is expanded along the first row using minors and cofactors.

Explanation:

We need to evaluate

\begin{vmatrix} 1! & 2! & 3! \\ 2! & 3! & 4! \\ 3! & 4! & 5! \end{vmatrix}

Expand along the first row:

= 1! \cdot (3! \cdot 5! - 4! \cdot 4!) - 2! \cdot (2! \cdot 5! - 4! \cdot 3!) + 3! \cdot (2! \cdot 4! - 3! \cdot 3!)

Compute each factorial:

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

Substitute and simplify:

= 1 \cdot (6 \cdot 120 - 24 \cdot 24) - 2 \cdot (2 \cdot 120 - 24 \cdot 6) + 6 \cdot (2 \cdot 24 - 6 \cdot 6)

= 1 \cdot (720 - 576) - 2 \cdot (240 - 144) + 6 \cdot (48 - 36)

= 1 \cdot 144 - 2 \cdot 96 + 6 \cdot 12

= 144 - 192 + 72 = 24

Answer:

The value of the determinant is

24

. Therefore, option C is correct.

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