1
$\begingroup$

$f(n)=β^n+α^n$ then $$ \begin{vmatrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \\ \end{vmatrix} $$

I don't quite understand which property of determinant can we use here? i tried using linear property but i did not get the required answer.

$\endgroup$
0
$\begingroup$

You just do not need to calculate $f(n)$ where $n$ is $1,2,3,4$ and substitute it in the matrix?

$f(1)$ for example should be $\beta + \alpha$

$\endgroup$
  • $\begingroup$ We have here a simmetric matrix, it can help? $\endgroup$ – Osvaldo Paniccia Sep 12 '16 at 19:46
  • $\begingroup$ I dont think there is any special property to calculate determinant of symmetric matrices. is there any? $\endgroup$ – Parul Sep 12 '16 at 19:52
  • $\begingroup$ Substituting doesn't give me the answer. Answer is βα(1-α)^2 * (1-β)^2 * (-β+α)^2. $\endgroup$ – Parul Sep 12 '16 at 19:55
  • $\begingroup$ No special rules, just do it following the calssic way and at the end, if im not wrong, you will have the expected result. $\endgroup$ – Osvaldo Paniccia Sep 12 '16 at 19:59
0
$\begingroup$

There is a trick, indeed. The values defined by $f(n)=\alpha^n+\beta^n$ obey the recurrence relation $$ f(n+2) - (\alpha+\beta)\,f(n+1) + (\alpha\beta)\,f(n) = 0$$ hence the original determinant equals $$ (1-\alpha)(1-\beta)\det\begin{pmatrix}1+f(0) & 1+f(1) & 1+f(2) \\ 1+f(1) & 1 + f(2) & 1+f(3) \\ 1 & 1 & 1\end{pmatrix}$$ or $$ (1-\alpha)(1-\beta)\det\begin{pmatrix}f(0) & f(1) & f(2) \\ f(1) & f(2) & f(3) \\ 1 & 1 & 1\end{pmatrix}$$ or $$ (1-\alpha)^2(1-\beta)^2\det\begin{pmatrix}f(0) & f(1) \\ f(1) & f(2)\end{pmatrix}$$ or $$ \boxed{\det\begin{pmatrix}1+f(0) & 1+f(1) & 1+f(2) \\ 1+f(1) & 1 + f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{pmatrix}=\color{red}{(1-\alpha)^2(1-\beta)^2 (\alpha-\beta)^2}.}\tag{1}$$


The properties exploited are

  • $\det\begin{pmatrix} r_1 \\ r_2 \\ r_3\end{pmatrix}=\det\begin{pmatrix} r_1 \\ r_2 \\ r_3-j r_2+k r_1\end{pmatrix}=\det\begin{pmatrix}r_1-r_3 \\ r_2-r_3 \\ r_3\end{pmatrix}$
  • $\det\begin{pmatrix} r_1 \\ r_2 \\ \alpha r_3\end{pmatrix}=\alpha\det\begin{pmatrix} r_1 \\ r_2 \\ r_3\end{pmatrix}$
  • $\det M^T = \det M$
  • $\det\begin{pmatrix}a & b & 0 \\ c & d & 0 \\ e & f & 1\end{pmatrix}=\det\begin{pmatrix}a & b \\ c & d\end{pmatrix}.$

Another tricky approach may be the following one. One may notice that the determinant is a polynomial in $\mathbb{Z}[\alpha,\beta]$ with degree $4+2=6$, and that the given matrix has rank one if $\alpha=1,\beta=1$ or $\alpha-\beta$. It follows that the determinant is an integer multiple of $(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$ and by performing an explicit computation at $\alpha=2,\beta=3$ the claim $(1)$ follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.