Matrix and Fibonacci number

Algebra Level 3

$F_n$ is the $n$ th Fibonacci number, defined by the recurrence relation $F_n=F_{n-1}+F_{n-2}$ with $F_1=F_2=1$ . If $n$ is a perfect square and $n >4$ , then

$\large \begin{vmatrix} F_{1}&F_{2}&\cdots&F_{\sqrt{n}} \\ F_{\sqrt{n}+1}&F_{\sqrt{n}+2}&\cdots&F_{2\sqrt{n}} \\ \vdots&\vdots&&\vdots \\ F_{n-\sqrt{n}+1}&F_{n-\sqrt{n}+2}&\cdots&F_{n} \end{vmatrix} = \ ?$

The answer is 0.

1 solution

Tommy Li
Oct 2, 2017

$\large \begin{vmatrix} F_{1}&F_{2}&\cdots&F_{\sqrt{n}} \\ F_{\sqrt{n}+1}&F_{\sqrt{n}+2}&\cdots&F_{2\sqrt{n}} \\ \vdots&\vdots&&\vdots \\ F_{n-\sqrt{n}+1}&F_{n-\sqrt{n}+2}&\cdots&F_{n} \end{vmatrix}$

$= \begin{vmatrix} F_{1}&F_{2}&F_{3}-F_{2}-F_{1}&\cdots&F_{\sqrt{n}} \\ F_{\sqrt{n}+1}&F_{\sqrt{n}+2}&F_{\sqrt{n}+3}-F_{\sqrt{n}+2}-F_{\sqrt{n}+1}&\cdots&F_{2\sqrt{n}} \\ \vdots&\vdots&\vdots&&\vdots \\ F_{n-\sqrt{n}+1}&F_{n-\sqrt{n}+2}& F_{n-\sqrt{n}+3}- F_{n-\sqrt{n}+2}- F_{n-\sqrt{n}+1}&\cdots&F_{n} \end{vmatrix}$

$= \begin{vmatrix} F_{1}&F_{2}&0&\cdots&F_{\sqrt{n}} \\ F_{\sqrt{n}+1}&F_{\sqrt{n}+2}&0&\cdots&F_{2\sqrt{n}} \\ \vdots&\vdots&\vdots&&\vdots \\ F_{n-\sqrt{n}+1}&F_{n-\sqrt{n}+2}&0&\cdots&F_{n} \end{vmatrix}$

$= 0$

Matrix and Fibonacci number

The answer is 0.

1 solution

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