Let $p, q, r$ be nonzero real numbers that are, respectively, the $10^{\text {th }}, 100^{\text {th }}$ and $1000^{\text {th }}$ terms of a harmonic progression. Consider the system of linear equations $$x+y+z=1$$ $$10 x+100 y+1000 z=0$$ $$q r x+p r y+p q z=0.$$ | List-I | | List-II | | |---|---|---|---| | (I) | If $\frac{q}{r}=10$, then the system of linear equations has | (P) | $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution | | (II) | If $\frac{p}{r} \neq 100$, then the system of linear equations has | (Q) | $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution | | (III) | If $\frac{p}{q} \neq 10$, then the system of linear equations has | (R) | infinitely many solutions | | (IV) | If $\frac{p}{q}=10$, then the system of linear equations has | (S) | no solution | | | | (T) | at least one solution | The correct option is:

(B) (I) → (Q); (II) → (S); (III) → (S); (IV) → (R)

(A) (I) → (T); (II) → (R); (III) → (S); (IV) → (T)

(C) (I) → (Q); (II) → (R); (III) → (P); (IV) → (R)

(D) (I) → (T); (II) → (S); (III) → (P); (IV) → (T)

Matrices — JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 Question with Solution

Let

p, q, r

be nonzero real numbers that are, respectively, the

1 0^{th}, 10 0^{th}

and

100 0^{th}

terms of a harmonic progression. Consider the system of linear equations

x + y + z = 1

10 x + 100 y + 1000 z = 0

q r x + p r y + pq z = 0.

List-I		List-II
(I)	If $\frac{q}{r} = 10$ , then the system of linear equations has	(P)	$x = 0, y = \frac{10}{9}, z = - \frac{1}{9}$ as a solution
(II)	If $\frac{p}{r} \neq = 100$ , then the system of linear equations has	(Q)	$x = \frac{10}{9}, y = - \frac{1}{9}, z = 0$ as a solution
(III)	If $\frac{p}{q} \neq = 10$ , then the system of linear equations has	(R)	infinitely many solutions
(IV)	If $\frac{p}{q} = 10$ , then the system of linear equations has	(S)	no solution
		(T)	at least one solution

The correct option is:

Answer: (B)

Since

p, q, r

are the

10

th,

100

th and

1000

th terms of a harmonic progression,

\frac{1}{p}, \frac{1}{q}, \frac{1}{r}

are in arithmetic progression, so

\frac{1}{r} - \frac{1}{q} = 10 (\frac{1}{q} - \frac{1}{p})

. Writing

a = \frac{1}{p}, b = \frac{1}{q}, c = \frac{1}{r}

, the coefficient determinant

11 a 110 b 1100 c = 9 (c - 11 b + 10 a)

vanishes identically, because the progression forces

c = 11 b - 10 a

; the system is therefore never uniquely solvable. Writing the third row as

λ (1, 1, 1) + μ (1, 10, 100)

fixes

μ = \frac{b - a}{9}

and

λ = \frac{10 a - b}{9}

, and the system is consistent (infinitely many solutions) exactly when

λ = 0

, i.e.

b = 10 a

, which means

\frac{p}{q} = 10

; otherwise it has no solution. Now

\frac{q}{r} = 10

gives

b = 10 a

, so there are infinitely many solutions and

x = \frac{10}{9}, y = - \frac{1}{9}, z = 0

is one of them. Each of

\frac{p}{r} \neq = 100

and

\frac{p}{q} \neq = 10

means

b \neq = 10 a

, giving no solution, while

\frac{p}{q} = 10

gives infinitely many solutions.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Single correct option