For α,β∈R and a natural number n, let Ar=r12n2+α2r2n2−β3r−232n(3n−1). Then 2A10−A8 is
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Official solution
We are given a determinant Ar defined as:
Ar=r12n2+α2r2n2−β3r−232n(3n−1)
We need to find the value of 2A10−A8.
Let's calculate the determinant Ar. We can use properties of determinants or expand it directly. A helpful trick is to use row operations to simplify the determinant before expanding. Let's perform the following row operations:
Replace Row 2 with (Row 2) - 2 × (Row 1)
Replace Row 3 with (Row 3) - 3 × (Row 1)
The determinant becomes:
Ar=r12n2+α2r−2r2−2(1)(n2−β)−2(2n2+α)(3r−2)−3r3−3(1)2n(3n−1)−3(2n2+α)
Let's simplify the entries in the new rows:
Row 2:
First element: 2r−2r=0
Second element: 2−2=0
Third element: (n2−β)−2(2n2+α)=n2−β−n2−2α=−2α−β
Row 3:
First element: (3r−2)−3r=−2
Second element: 3−3=0
Third element: 2n(3n−1)−3(2n2+α)=23n2−n−23n2−3α=23n2−n−3n2−3α=−2n−3α
So the determinant simplifies to:
Ar=r12n2+α00−2α−β−20−2n−3α
Now, we can easily calculate this determinant by expanding along the second column, since it has two zeros. The expansion along the second column is:
Ar=−1⋅∣0−2−2α−β−2n−3α∣Ar=−1⋅((0)×(−2n−3α)−(−2)×(−2α−β))Ar=−1⋅(0−(4α+2β))Ar=−1⋅(−4α−2β)Ar=4α+2β
Notice that the value of the determinant Ar does not depend on r at all! It's a constant value determined by α and β.
Now we need to find the value of 2A10−A8.
Since Ar=4α+2β for any value of r, we have:
A10=4α+2βA8=4α+2β
So, 2A10−A8=2(4α+2β)−(4α+2β)2A10−A8=8α+4β−4α−2β2A10−A8=(8α−4α)+(4β−2β)2A10−A8=4α+2β
The value of the expression 2A10−A8 is 4α+2β.