For , R and a natural number n, let Ar = vmatrix r & 1 & n22 + \ 2r & 2 & n2 - \ 3r - 2 & 3 & n(3n - 1)2 vmatrix. Then 2A10 - A8 is
- 4 + 2
- 0
- 2n
- 2 + 4
Answer
(1) 4 + 2
Solution
Only the first column of Ar depends on r, and a determinant is linear in a column, so 2A10 - A8 is a single determinant with C1 = 2(10, 20, 28) - (8, 16, 22) = (12, 24, 34):
2A10 - A8 = vmatrix 12 & 1 & n22 + \ 24 & 2 & n2 - \ 34 & 3 & n(3n - 1)2 vmatrix
Applying R2 R2 - 2R1 and R3 R3 - 3R1 gives
2A10 - A8 = vmatrix 12 & 1 & n22 + \ 0 & 0 & -2 - \ -2 & 0 & - n2 - 3 vmatrix
Expanding along R2: 2A10 - A8 = -(-2 - ) vmatrix 12 & 1 \ -2 & 0 vmatrix = (2 + ) 2 = 4 + 2
The value is free of n, and 0, 2n, 2 + 4 do not match it, so (2), (3) and (4) are wrong.
Answer: (1)