Solveeit Logo
Login

Question

Question: If p, q, r are in A.P., then the determinant \(\left| \begin{matrix} a^{2} + a^{2n + 1} + 2p & b^{2...

If p, q, r are in A.P., then the determinant

∣a2+a2n+1+2pb2+2n+2+3qc2+p2n+p2n+1+q2qa2+2n+pb2+2n+1+2qc2−r∣\left| \begin{matrix} a^{2} + a^{2n + 1} + 2p & b^{2} + 2^{n + 2} + 3q & c^{2} + p \\ 2^{n} + p & 2^{n + 1} + q & 2q \\ a^{2} + 2^{n} + p & b^{2} + 2^{n + 1} + 2q & c^{2} - r \end{matrix} \right|=

A

1

B

0

C

a2b2c2 – 2n

D

(a2 + b2 + c2) – 2n q

Answer

0

Explanation

Solution

The given determinant

= ∣2n+1−2n+p2n+2−2n+1+qp+r2n+p2n+1+qp+ra2+2n+pb2+2n+1+2qc2−r∣\left| \begin{matrix} 2^{n + 1} - 2^{n} + p & 2^{n + 2} - 2^{n + 1} + q & p + r \\ 2^{n} + p & 2^{n + 1} + q & p + r \\ a^{2} + 2^{n} + p & b^{2} + 2^{n + 1} + 2q & c^{2} - r \end{matrix} \right| (using R1 → R1 – R3 and 2q = p + r) = ∣2n(2−1)+p2n+1(2−1)+qp+r2n+p2n+1+qp+ra2+2n+pb2+2n+1+2qc2−r∣\left| \begin{matrix} 2^{n}(2 - 1) + p & 2^{n + 1}(2 - 1) + q & p + r \\ 2^{n} + p & 2^{n + 1} + q & p + r \\ a^{2} + 2^{n} + p & b^{2} + 2^{n + 1} + 2q & c^{2} - r \end{matrix} \right|

= ∣2n+p2n+1+qp+r2n+p2n+1+qp+ra2+2n+pb2+2n−1+2qc2−r∣\left| \begin{matrix} 2^{n} + p & 2^{n + 1} + q & p + r \\ 2^{n} + p & 2^{n + 1} + q & p + r \\ a^{2} + 2^{n} + p & b^{2} + 2^{n - 1} + 2q & c^{2} - r \end{matrix} \right| = 0

(Q R1 ≡ R2)