Determinants JEE
∣ ∣ ∣ ∣ ∣ ∣ x n y n z n x n + 2 y n + 2 z n + 2 x n + 3 y n + 3 z n + 3 ∣ ∣ ∣ ∣ ∣ ∣ = ( x − y ) ( y − z ) ( z − x ) ( x 1 + y 1 + z 1 ) , then n equals
Also see Coordinate Geometry
The answer is -1.
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3 solutions
JEE Approach : Put x=4,y=2,z=1 , n comes out to be -1. You can take any other values of x,y,z too, but not 0. Also try to avoid to take those values which will make R.H.S. equal to zero.
Hi Saketh , I'm not sure if I could completely understand your solution , can you be a little bit more descriptive ?
Your solution is quite good for JEE and I'm interested to know about it :)
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Yes, see you can see that in determinant, if you think about expanding along the first row, you will note that the x terms only multiply with y or z and not amongst themselves, so the highest power if x is the same as what you can see in the determinant namely n+3,
On the right, the highest power of x as you can see is 2 (choose x from 1st and 3rd bracket, then chose 1/y or 1/z from last bracket to get power as 2)
these must be equal,
so n+3 = 2 or n = -1
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Oh , I get it now. So observation skills are enough to solve this question :) +1
Did exactly same
Same approach!
After solving the determinant by taking out x n , y n , z n common from the columns , we are left with x n + 1 ⋅ y n + 1 ⋅ z n + 1 ⋅ ( x − y ) ( y − z ) ( z − x ) ( x 1 + y 1 + z 1 )
Now we note that there are no terms of the form x n + 1 ⋅ y n + 1 ⋅ z n + 1 in the final answer which means their exponents are equal to 0 .
∴ n + 1 = 0 ⇒ n = − 1
If we consider the degree of the matrix it is 3n+5.And the degree of the equation on RHS is 2.On equating both we get x=-1
Max power of 2 in determinant = n+3 (as all x are in same row, they cant multiply with each other which can be easily seen without expanding)
max power on right side is 2, (when you jump from x in 1st bracket to x in 3rd and finally to a y or z)
so n+3=2 or n =-1