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Let $n$ be a positive integer and

\begin{align} M = \begin{pmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! &(n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \\ \end{pmatrix} \end{align}

Use properties of determinant to show that \begin{align}\left(\frac{|M|}{(n!)^3}- 4\right)\end{align} is divisible by $n$.

I took an $n!$ factor out of the matrix, getting a new matrix $B$ such that $\det A = n^3 \det B$, since $A$ and $B$ are row equivalents. But what is $\det B$ ?

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After you take the common factor $n!$ out of all columns, take $(n+1)$ and $(n+1)(n+2)$ out of 2nd and 3rd columns, respectively: $$\frac{|M|}{(n!)^3}=\left|\begin{array}{ccc} 1 & \color{red}{n+1} & \color{blue}{(n+1)(n+2)}\\ n+1 & (\color{red}{n+1})(n+2) & \small{\color{blue}{(n+1)(n+2)}(n+3)}\\ (n+1)(n+2) & \small{(\color{red}{n+1})(n+2)(n+3)} & \scriptsize{\color{blue}{(n+1)(n+2)}(n+3)(n+4)} \end{array}\right|=\\ \color{red}{(n+1)}\color{blue}{(n+1)(n+2)}\left|\begin{array}{ccc} 1 & 1 & 1\\ n+1 & n+2 & \small{n+3}\\ (n+1)(n+2) & \small{(n+2)(n+3)} & \scriptsize{(n+3)(n+4)} \end{array}\right|\stackrel{C_3-C_2\to C_3\\ C_2-C_1\to C_2}{=}\\ (n+1)^2(n+2)\left|\begin{array}{ccc} 1 & 0 & 0\\ n+1 & 1 & 1\\ (n+1)(n+2) & 2n & 2n+2 \end{array}\right|\stackrel{C_3-C_2\to C_3}=\\ (n+1)^2(n+2)\left|\begin{array}{ccc} 1 & 0 & 0\\ n+1 & 1 & 0\\ (n+1)(n+2) & 2n & \color{red}2 \end{array}\right|=\color{red}2(n+1)^2(n+2)$$ Hence: $$\begin{align}\left(\frac{|M|}{(n!)^3}- 4\right)\end{align}=2n^3+8n^2+10n\equiv 0 \pmod{n}.$$

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  • $\begingroup$ You are welcome. Thank you all for understanding. Good luck! $\endgroup$ – farruhota Aug 20 '19 at 9:37
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So, they're looking for the determinant of $M$. We can make that easier to calculate by factoring a common term of $n!$ out of each row, giving us a new matrix:

$$N=\left[\begin{array}{ccc} 1 & n+1 & (n+1)(n+2)\\ n+1 & (n+1)(n+2) & \small{(n+1)(n+2)(n+3)}\\ (n+1)(n+2) & \small{(n+1)(n+2)(n+3)} & \scriptsize{(n+1)(n+2)(n+3)(n+4)} \end{array}\right]$$

Now, $|N|=\frac{|M|}{(n!)^3}$, so we just need to find $|N|-4$. It looks like we're about to dive into horrible polynomial multiplications. But wait! We only need to show that $|N|-4$ is divisible by $n$. That means that if the constant term of the polynomial $|N|$ is 4, we're done. We can do that by multiplying those terms out and ignoring all but the constant term"

$$\left[\begin{array}{ccc} 1 & ...+1 & ...+2\\ ..+1 & ...+2 & ...+6\\ ...+2 & ...+6 & ...+24 \end{array}\right]$$

Now the determinant is easy enough to calculate. Using Sarrus' Rule, the value of this determinant is $$(1\cdot2\cdot24) + (1\cdot6\cdot2) +(2\cdot1\cdot6)- (2\cdot2\cdot2) - (1\cdot6\cdot6) - (1\cdot1\cdot24) $$ $$ = 48+12+12-8-36-24=4$$.

Therefore, $|N|=An^6+Bn^5+Cn^4+Dn^3+En^2+Fn+4$ for some integer values of $A,B,C,D,E,F$. We don't need to find those values, though. The only thing that matters is that every term of $|N|-4$ contains a power of $n$, which is what we were asked to show.

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  • $\begingroup$ how the determinant could be 4 i don't understand $\endgroup$ – Anupa Kulathunga Aug 20 '19 at 7:30
  • $\begingroup$ That should be clearer. The determinant of that last matrix is 4, which shows that the determinant on N is a polynomial of the form $n^6+An^5+...+Fn+4$, which proves the statement that $|N|-4$ is divisible by $n$. $\endgroup$ – Matthew Daly Aug 20 '19 at 7:33
  • $\begingroup$ don't understand yet] $\endgroup$ – Anupa Kulathunga Aug 20 '19 at 7:42
  • $\begingroup$ How far do you get before not understanding? $\endgroup$ – Matthew Daly Aug 20 '19 at 7:45
  • $\begingroup$ I understand how to collect the scaler values of the matrix elements and but I didn't understand how you get the 4 value by them $\endgroup$ – Anupa Kulathunga Aug 20 '19 at 7:49

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