# find value of determinant

$f(n)=β^n+α^n$ 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.

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 $\beta + \alpha$

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

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.