Q151 / 270 · NDA 2022 (Phase 1) · Mathematics
If $\quad \Delta_1=\left|\begin{array}{lll}1 & p & q \\ 1 & q & r \\ 1 & r & p\end{array}\right|$ and $\Delta_2=\left|\begin{array}{lll}1 & 1 & 1 \\ q & r & p \\ r & p & q\end{array}\right| \quad$ where $p \neq q \neq r$, then $\Delta_1+\Delta_2$ is
- A0
- Balways positive
- Correct answerCalways negative
- Dpositive if $p, q, r$ are positive else negative
Solution
$\begin{aligned}
& \because \Delta_1^1=\left|\begin{array}{lll}
1 & 1 & 1 \\
p & q & r \\
q & r & p
\end{array}\right|=\left|\begin{array}{lll}
1 & 1 & 1 \\
q & r & p \\
r & p & q
\end{array}\right|\left[\begin{array}{l}
\text { Applying } \\
c_1 \longleftrightarrow c_2 \\
c_2 \longleftrightarrow c_3
\end{array}\right] \\
& \therefore \quad \Delta_1^1=\Delta_1=\Delta_2 \\
& \therefore \quad \Delta_1+\Delta_2=2 \Delta_2=2\left|\begin{array}{lll}
1 & 1 & 1 \\
q & r & p \\
r & p & q
\end{array}\right|
\end{aligned}$
[Applying $c_2 \rightarrow c_2-c_1$ and $c_3 \rightarrow c_3-c_1$ ]
$\begin{aligned} & \because \Delta_1^1=\left|\begin{array}{lll}1 & 1 & 1 \\ p & q & r \\ q & r & p\end{array}\right|=\left|\begin{array}{lll}1 & 1 & 1 \\ q & r & p \\ r & p & q\end{array}\right|\left[\begin{array}{l}\text { Applying } \\ c_1 \longleftrightarrow c_2 \\ c_2 \longleftrightarrow c_3\end{array}\right] \\ & \therefore \quad \Delta_1^1=\Delta_1=\Delta_2 \\ & \therefore \quad \Delta_1+\Delta_2=2 \Delta_2=2\left|\begin{array}{lll}1 & 1 & 1 \\ q & r & p \\ r & p & q\end{array}\right|\end{aligned}$
[Applying $c_2 \rightarrow c_2-c_1$ and $c_3 \rightarrow c_3-c_1$ ]
$=2\left|\begin{array}{ccc}1 & 0 & 0 \\ q & r-q & p-q \\ r & p-r & q-r\end{array}\right|$
$\begin{aligned} & =-2\left[p^2+q^2+r^2-p q-q r-p r\right] \\ & =-\left[(p-q)^2+(q-r)^2+(r-p)^2\right] \lt 0\end{aligned}$
Always negative.
Hence, the correct answer is option: 'C'.