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CGPET 2022 Solved Paper

Section: Mathematics

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Question : 137 of 150

Marks: +1, -0

If

D=\begin{vmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{vmatrix}

\;\frac{D}{2(5!)^3}-2

a multiple of 5
a multiple of 6
0
$-5$

Solution:

We have,

D=\begin{vmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{vmatrix}

=\begin{vmatrix} 5! & 6 \times 5! & 7 \times 6 \times 5! \\ 6 \times 5! & 7 \times 6 \times 5! & 8 \times 7 \times 6 \times 5! \\ 7 \times 6 \times 5! & 8 \times 7 \times 6 \times 5! & 9 \times 8 \times 7 \times 6 \times 5! \end{vmatrix}

=(5!)^3\begin{vmatrix} 1 & 6 & 42 \\ 6 & 42 & 336 \\ 42 & 336 & 3024 \end{vmatrix}

Common 6 from

R_2

and

42

from

R_3

,

(5!)^3 \times 6 \times 42\begin{vmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{vmatrix}

On applying

R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_2

, we get

(5!)^3 \times 6 \times 42\begin{vmatrix} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{vmatrix}

=(5!)^3 \times 6 \times 42 \times 1[16-14]

D=(5!)^3 \times 6 \times 42 \times 2

\Rightarrow \frac{D}{2(5!)^3} = 6 \times 42 = 252

\therefore \frac{D}{2(5!)^3} - 2 = 250

which is a multiple of 5 .

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