SATHEE: Matrices and Determinants 2 Question 36

Matrices and Determinants 2 Question 36

40. For a fixed positive integer $n$ , if

$D = | \begin{matrix} n! & (n + 1)! & (n + 2)! \\ (n + 1)! & (n + 2)! & (n + 3)! \\ (n + 2)! & (n + 3)! & (n + 4)! \end{matrix} |$

then show that $\frac{D}{(n!)^{3}} - 4$ is divisible by $n . (1992, 4 M)$

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Solution:

Given, $D = | \begin{matrix} n! & (n + 1)! & (n + 2)! \\ (n + 1)! & (n + 2)! & (n + 3)! \\ (n + 2)! & (n + 3)! & (n + 4)! \end{matrix} |$

Taking $n$ !, $(n + 1)$ ! and $(n + 2)$ ! common from $R_{1}, R_{2}$ and $R_{3}$ , respectively.

$∴ D = n! (n + 1)! (n + 2)! | \begin{matrix} 1 & (n + 1) & (n + 1) (n + 2) \\ 1 & (n + 2) & (n + 2) (n + 3) \\ 1 & (n + 3) & (n + 3) (n + 4) \end{matrix} |$

Applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{2}$ , we get

$D = n! (n + 1)! (n + 2)! | \begin{matrix} 1 & (n + 1) & (n + 1) (n + 2) \\ 0 & 1 & 2 n + 4 \\ 0 & 1 & 2 n + 6 \end{matrix} |$

Expanding along $C_{1}$ , we get

$\begin{aligned} D = (n!) (n + 1)! (n + 2)! [(2 n + 6) - (2 n + 4)] \\ D = (n!) (n + 1)! (n + 2)! [2] \end{aligned}$

On dividing both side by $(n!)^{3}$

$\begin{aligned} \Rightarrow \frac{D}{(n!)^{3}} = \frac{(n!) (n!) (n + 1) (n!) (n + 1) (n + 2) 2}{(n!)^{3}} \\ \Rightarrow \frac{D}{(n!)^{3}} = 2 (n + 1) (n + 1) (n + 2) \\ \Rightarrow \frac{D}{(n!)^{3}} = 2 (n^{3} + 4 n^{2} + 5 n + 2) = 2 n (n^{2} + 4 n + 5) + 4 \\ \Rightarrow \frac{D}{(n!)^{3}} - 4 = 2 n (n^{2} + 4 n + 5) \end{aligned}$

which shows that $\frac{D}{(n!)^{3}} - 4$ is divisible by $n$ .

Matrices and Determinants 2 Question 36

40. For a fixed positive integer $n$ , if

Solution:

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Matrices and Determinants 2 Question 36

40. For a fixed positive integer n, if

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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40. For a fixed positive integer $n$ , if