Think of it the other way!
Consider the following system of linear equations:
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 1 3 2 2 4 3 1 0 2 4 3 1 2 5 + + + + + a 1 6 a 8 1 a 2 5 6 a 6 2 5 a + + + + + b 8 b 2 7 b 6 4 b 1 2 5 b + + + + + c 4 c 9 c 1 6 c 2 5 c + + + + + d 2 d 3 d 4 d 5 d + + + + + e e e e e = = = = = 1 2 3 4 5 .
Evaluate
∣ ∣ ∣ ∣ 1 2 5 a e 8 b c d ∣ ∣ ∣ ∣ .
The answer is 187.
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2 solutions
Discussions for this problem are now closed
Another way is to subtract two consecutive equations and reduce the case equations in each step from five equations to four, then three, then two and finally one equation (which will be a linear equation in one variable). We then substitute known values of variables obtained by working reversely and ultimately getting the values of all unknown variables. The rest is child's play. :D
It is one of the possible functional equation for G(x). Can you prove that it is only possible equation. It is nowhere mentioned that g(x) is a polynomial function. (However, I solved it by taking polynomial function and only then answer matches.)
Suppose we have a monic degree 5 function f ( x ) such that f ( k ) = k for k = 1 → 5 (as we have in this scenario).
Let g ( x ) = f ( x ) − x . This means that g ( k ) = 0 for k = 1 → 5 . Because g ( x ) is degree 5 and monic, it must be that g ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 )
Thus, f ( x ) = g ( x ) + x = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) + x
I'm not saying that you are wrong. Instead, I am saying that it is one of the many functions that satisfy given condition. What happens if you think that it is not a monic fifth degree polynomial function.
May be we get a function of the form e^ax+tan(x)+........ which satisfy above given condition. However, by seeing the coefficients one can guess that it is monic polynomial function of degree 5.
@Prakash Chandra Rai – We are assuming it is a polynomial because when substituting x = 1 , 2 , 3 , 4 , 5 , it gives us our system of equations. We are using it as a kind of "bridge" to arrive at the solution ( a , b , c , d , e ) . Of course, it might not be a monic degree 5 polynomial that satisfies f ( k ) = k for k = 1 → 5 , but if it isn't then it is useless to us. Sure, we can say that f ( x ) = x works, but that doesn't tell us anything about a , b , c , d , e .
@Daniel Liu – I agree. @PrakashChandraRai, we are just using a function of our comfort to determine and prove the uniqueness of the solutions. The polynomial function is just a "bridge" to be used to our advantage. And @AareyanManzoor is right. Suppose, there was another set of non-polynomial relation that was dug out to fit the relations. But, then invoke your knowledge of the uniqueness as we just concluded (a,b,c,d,e were concluded as fixed) or use the uniqueness inherent in the set of five linear equations to chart it out: We conclude: Even that method would yield identical answers.
@Prakash Chandra Rai – a , b , c , d , e are in the 1 s t d e g r e e and there are 5 equations, so there should be 1 pair of solution.if you keep adding and subtracting the equations, you'll get the solution, but this is the easier way.
For those who didn't understand Prakash Chandra Rai comment (answer below):
This solution takes the form of: Let g(x) be a polynomial such that ... then g(x) coefficients satisfy those equations for a,b,c,d,e. It then proceeds to calculate g(x) coefficients from a different perspective.
That is certainly a method for generating a solution, but the question whether there are other solutions remains open. The solution is not of the form: Let a,b,c,d,e be one solution of this system then there is a polynomial with coefficients ...
So there could be other solutions that have nothing to do with any polynnomials. Well, there isn't. The answer why:
The idea is what Aareyan Manzoor said below, a system of 5 equations and 5 variables should have unique solution. Obviously, that is not always the case, so the proof is to observe that the coefficients of this linear system form a Vandermonde matrix . This is a known matrix, and it's determinant is easy to calculate, but suffices to say that it is non zero. Hope you know some linear algebra to understand why this solves the problem.
Cheers.
Wow, great solution! I hadn't notice it when I first saw, but yeah, it's quite easy to realize it when you think as a polynomial. I solved it other way, by subtracting lines, not as brilliant as your solution, though.
Congratulations on you nice approach.
Specifically best.
We begin to reduce the number of equations in each step by subtracting first equation from second, second from third, and so on. In short, we subtract the k th equation from the ( k + 1 ) th equation to reduce number of equations in each successive step.
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ 1 + a + b + c + d + e = 1 3 2 + 1 6 a + 8 b + 4 c + 2 d + e = 2 2 4 3 + 8 1 a + 2 7 b + 9 c + 3 d + e = 2 1 0 2 4 + 2 5 6 a + 6 4 b + 1 6 c + 4 d + e = 4 3 1 2 5 + 6 2 5 a + 1 2 5 b + 2 5 c + 5 d + e = 5 ⟹ ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 3 1 + 1 5 a + 7 b + 3 c + d = 1 2 1 1 + 6 5 a + 1 9 b + 5 c + d = 1 7 8 1 + 1 7 5 a + 3 7 b + 7 c + d = 1 2 1 0 1 + 3 6 9 a + 6 1 b + 9 c + d = 1 ⟹ ⎩ ⎪ ⎨ ⎪ ⎧ 1 8 0 + 5 0 a + 1 2 b + 2 c = 0 5 7 0 + 1 1 0 a + 1 8 b + 2 c = 0 1 3 2 0 + 1 9 4 a + 2 4 b + 2 c = 0 ⟹ { 3 9 0 + 6 0 a + 6 b = 0 7 5 0 + 8 4 a + 6 b = 0 ⟹ 3 6 0 + 2 4 a = 0 ⟹ a = ( − 1 5 )
Working reversely by substituting values in equations of above cases, we can get the other values, b = 8 5 , c = ( − 2 2 5 ) , d = 2 7 5 , e = ( − 1 2 0 ) .
Plugging the values in the required expression gives us the answer 1 8 7
The other approach to this problem is ,I guess,the intended one,but yours is also essential to know,especially the systematic way in which you handled the equations is admirable.
think of it as a polynomial, let there be a polynomial G ( x ) = x 5 + a x 4 + b x 3 + c x 2 + d x + e it fits all the symptoms as G ( 1 ) , G ( 2 ) , G ( 3 ) , G ( 4 ) , G ( 5 ) ,so we get G ( n ) = n f o r a l l n ∈ [ 1 , 2 , 3 , 4 , 5 ] we can write G ( x ) as: G ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) + x it satisfies all equations, expand and we get G ( x ) = x 5 − 1 5 x 4 + 8 5 x 3 − 2 2 5 x 2 + 2 7 5 x − 1 2 0 and ∣ ∣ ∣ ∣ 1 2 5 × − 1 5 × − 1 2 0 8 × 8 5 × − 2 2 5 × 2 7 5 ∣ ∣ ∣ ∣ = 1 8 7