JEE Advanced 2022 Paper 1 Q17 Mathematics Matrices & Determinants Systems of Equations Hard

JEE Advanced 2022 Paper 1 · Q17 · Systems of Equations

Let $p,q,r$ be nonzero real numbers that are, respectively, the $10^{th}$, $100^{th}$ and $1000^{th}$ terms of a harmonic progression. Consider the system of linear equations $x+y+z=1$; $10x+100y+1000z=0$; $qr\,x+pr\,y+pq\,z=0$.

List-I: (I) If $q/r=10$, then the system of linear equations has (II) If $p/r\ne 100$, then the system of linear equations has (III) If $p/q\ne 10$, then the system of linear equations has (IV) If $p/q=10$, then the system of linear equations has

List-II: (P) $x=0, y=10/9, z=-1/9$ as a solution (Q) $x=10/9, y=-1/9, z=0$ as a solution (R) infinitely many solutions (S) no solution (T) at least one solution

The correct option is:

  1. A. (I)$\to$(T); (II)$\to$(R); (III)$\to$(S); (IV)$\to$(T)
  2. B. (I)$\to$(Q); (II)$\to$(S); (III)$\to$(S); (IV)$\to$(R)
  3. C. (I)$\to$(Q); (II)$\to$(R); (III)$\to$(P); (IV)$\to$(R)
  4. D. (I)$\to$(T); (II)$\to$(S); (III)$\to$(P); (IV)$\to$(T)
Reveal answer + step-by-step solution

Correct answer:B

Solution

If $1/p,1/q,1/r$ are AP, write $1/p=a+9d, 1/q=a+99d, 1/r=a+999d$ for some $a,d$. Multiply 3rd equation by $1/(pqr)$: $\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=0$, i.e. $(a+9d)x+(a+99d)y+(a+999d)z=0$. (I) $q/r=10\Rightarrow 1/r=10/q\Rightarrow a+999d=10(a+99d)\Rightarrow a+999d=10a+990d\Rightarrow 9d=9a\Rightarrow a=d$. Then $1/p=10a,1/q=100a,1/r=1000a$, giving $p=1/(10a)$ etc. The third equation becomes $10a\cdot x+100a\cdot y+1000a\cdot z=0$, identical (after dividing) to eq2. So system reduces to two equations in three unknowns — infinitely many solutions, but we need to check if Q's specific tuple $(10/9,-1/9,0)$ satisfies: $10/9-1/9+0=1$ ✓; $10\cdot 10/9+100\cdot(-1/9)=100/9-100/9=0$ ✓. So (I)$\to$Q. (II) $p/r\ne 100$: the 3rd equation is genuinely different from a multiple of eq2; system over-determined likely $\to$ no solution generically (S). (III) $p/q\ne 10$: similar, no solution (S). (IV) $p/q=10\Rightarrow 1/q=10/p\Rightarrow a+99d=10(a+9d)\Rightarrow a+99d=10a+90d\Rightarrow 9d=9a\cdot(?)$ — gives $9d=9a$ — wait $99d-90d=10a-a\Rightarrow 9d=9a\Rightarrow a=d$ — same as I, infinitely many (R). Pattern matches option (B).

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →