Determinant

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diogomgf

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Show that the [MATH]detF_{n} = detF_{n-1} + detF_{n-2} , n \geq 3[/MATH]
If we apply Laplace's Theorem on column 1 its easy to see that [MATH]detF_{n} = detF_{n}(1|1) + detF_{n}(2|1) = detF_{n-1} + detF_{n}(2|1) [/MATH]I just don't know how to show that [MATH]detF_{n}(2|1) = detF_{n-2}[/MATH].

Can some 1 help me?
 
Jomo said:
Oh, well. Please get going with induction and post back.
Click to expand...

This is my attemp at induction in this case (though I am not sure if this is the correct way to go about it):
Induction hypothesis: We assume [MATH]detF_{n} = detF_{n-1} + detF_{n-2}[/MATH] is true.
[MATH]detF_{n} = detF_{n-1} + detF_{n-2} \Rightarrow detF_{n+1} = detF_{n} + detF_{n-1}[/MATH].

Now:
-[MATH]detF_{n+1} =^{Lapl.}{c_{1}} detF_{n+1}(1|1) + detF_{n+1}(2|1)[/MATH].
-[MATH]detF_{n} =^{Lapl.}{c_{1}} detF_{n}(1|1) + detF_{n}(2|1)[/MATH].
-[MATH]detF_{n- 1} =^{Lapl.}{c_{1}} detF_{n-1}(1|1) + detF_{n-1}(2|1)[/MATH].
-[MATH]detF_{n- 2} =^{Lapl.}{c_{1}} detF_{n-2}(1|1) + detF_{n-2}(2|1)[/MATH].

So:
[MATH]detF_{n+1} = detF_{n} + detF_{n-1}= detF_{n-1}(1|1) + detF_{n-1}(2|1) + detF_{n-2}(1|1) + detF_{n-2}(2|1) + detF_{n-1}(1|1) + detF_{n-1}(2|1) = 2 \cdot detF_{n-1} + detF_{n-2}[/MATH]
This just feels like walking in circles...
 
Last edited:
Jomo said:
And the base case is??
Click to expand...

In the O.P I wrote [MATH]n \geq 3[/MATH] but actually I just double checked the exercise and it isn't there.

So the base case is:

[MATH]detF_{3} = detF_{2} + detF{1}[/MATH]:
[MATH]detF_{3}= 1 \cdot 1 \cdot 1 +1 \cdot 1 \cdot 0 + -1 \cdot -1 \cdot 0 - ( 1 \cdot 0 \cdot 1 + 1 \cdot 1 \cdot -1 + -1 \cdot 1 \cdot 1) = 1 - (-1-1) = 1+2 = 3[/MATH].
[MATH]detF_{2} = 1 \cdot 1 - ( -1 \cdot 1) = 1-(-1) = 2[/MATH].
[MATH]detF_{1} = 1[/MATH].
[MATH]detF_{3} = detF_{2} + detF{1} , 3 = 2 + 1[/MATH].
 
@Jomo

So this solves the problem? The induction method was as I presented above? I hate induction... Feels wrong all the time.
 
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