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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Determinants

NDA 2022 · Mathematics · Determinants

If Δ_1=| array lll 1 & p & q 1 & q & r 1 & r & p array | and Δ_2=|...

If \quad \Delta_1=\left|1pq1qr1rp\begin{array}{lll}1 & p & q \\ 1 & q & r \\ 1 & r & p\end{array}\right| and \Delta_2=\left|111qrprpq\begin{array}{lll}1 & 1 & 1 \\ q & r & p \\ r & p & q\end{array}\right| \quad where p≠q≠rp \neq q \neq r, then Δ1+Δ2\Delta_1+\Delta_2 is

Options

  1. A0
  2. Balways positive
  3. Calways negative
  4. Dpositive if p,q,rp, q, r are positive else negative

Correct answer

C. always negative

Step-by-step solution

∵Δ11=∣111pqrqrp∣=∣111qrprpq∣[ Applying c1⟷c2c2⟷c3]∴Δ11=Δ1=Δ2∴Δ1+Δ2=2Δ2=2∣111qrprpq∣\begin{aligned} & \because \Delta_1^1=\left|\begin{array}{lll} 1 & 1 & 1 \\ p & q & r \\ q & r & p \end{array}\right|=\left|\begin{array}{lll} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{array}\right|\left[\begin{array}{l} \text{ Applying } \\ c_1 \longleftrightarrow c_2 \\ c_2 \longleftrightarrow c_3 \end{array}\right] \\ & \therefore \quad \Delta_1^1=\Delta_1=\Delta_2 \\ & \therefore \quad \Delta_1+\Delta_2=2 \Delta_2=2\left|\begin{array}{lll} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{array}\right| \end{aligned} [Applying c2→c2−c1c_2 \rightarrow c_2-c_1 and c3→c3−c1c_3 \rightarrow c_3-c_1 ] ∵Δ11=∣111pqrqrp∣=∣111qrprpq∣[ Applying c1⟷c2c2⟷c3]∴Δ11=Δ1=Δ2∴Δ1+Δ2=2Δ2=2∣111qrprpq∣\begin{aligned} & \because \Delta_1^1=\left|\begin{array}{lll}1 & 1 & 1 \\ p & q & r \\ q & r & p\end{array}\right|=\left|\begin{array}{lll}1 & 1 & 1 \\ q & r & p \\ r & p & q\end{array}\right|\left[\begin{array}{l}\text{ Applying } \\ c_1 \longleftrightarrow c_2 \\ c_2 \longleftrightarrow c_3\end{array}\right] \\ & \therefore \quad \Delta_1^1=\Delta_1=\Delta_2 \\ & \therefore \quad \Delta_1+\Delta_2=2 \Delta_2=2\left|\begin{array}{lll}1 & 1 & 1 \\ q & r & p \\ r & p & q\end{array}\right|\end{aligned} [Applying c2→c2−c1c_2 \rightarrow c_2-c_1 and c3→c3−c1c_3 \rightarrow c_3-c_1 ] =2\left|100qr−qp−qrp−rq−r\begin{array}{ccc}1 & 0 & 0 \\ q & r-q & p-q \\ r & p-r & q-r\end{array}\right|=−2[p2+q2+r2−pq−qr−pr]=−[(p−q)2+(q−r)2+(r−p)2]<0\begin{aligned} & =-2\left[p^2+q^2+r^2-p q-q r-p r\right] \\ & =-\left[(p-q)^2+(q-r)^2+(r-p)^2\right] < 0\end{aligned}$$ Always negative.
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