# Use properties of determinant and show

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 $$\det B$$ ?

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}.

• You are welcome. Thank you all for understanding. Good luck! – farruhota Aug 20 '19 at 9:37

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.

• how the determinant could be 4 i don't understand – Anupa Kulathunga Aug 20 '19 at 7:30
• 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 $n$. – Matthew Daly Aug 20 '19 at 7:33
• don't understand yet] – Anupa Kulathunga Aug 20 '19 at 7:42
• How far do you get before not understanding? – Matthew Daly Aug 20 '19 at 7:45
• 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 – Anupa Kulathunga Aug 20 '19 at 7:49