Let p,q,r be non-zero real numbers that are respectively the 10th,100th and 1000th 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 Column-I Column-II(I)If qr=10, then the system of linear equations has(P)x=0,y=109,z=−19 as a solution(II)If pr≠100, then the system of linear equations has(Q)x=109,y=−19,z=0 as a solution(III)If pq≠10, then the system of linear equations has(R)infinitely many solutions(IV)If pq=10, then the system of linear equations has(S)no solution (T)at least one solutionThe correct option is:

(I) qr=10⇒D=Dx=Dy=Dz=0 So, there will be infinitely many solutions and they will be of the type (x,y,z)=(109+10λ,−19−11λ,λ) which gives (I)-(Q),(R),(T)(II) pr≠100⇒Dy≠0So, no solution which gives (II)-(S)(III) pq≠10⇒Dz≠0So, no solution which gives (III)-(S)(IV) pq=10⇒Dx=Dy=Dz=D=0So, infinitely many solutions which give (IV)-(Q),(R),(T)

Let p,q,r be non-zero real numbers that are respectively the 10th,100th and 1000th 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 Column-I Column-II(I)If qr=10, then the system of linear equations has(P)x=0,y=109,z=−19 as a solution(II)If pr≠100, then the system of linear equations has(Q)x=109,y=−19,z=0 as a solution(III)If pq≠10, then the system of linear equations has(R)infinitely many solutions(IV)If pq=10, then the system of linear equations has(S)no solution (T)at least one solutionThe correct option is:

(I) qr=10⇒D=Dx=Dy=Dz=0 So, there will be infinitely many solutions and they will be of the type (x,y,z)=(109+10λ,−19−11λ,λ) which gives (I)-(Q),(R),(T)(II) pr≠100⇒Dy≠0So, no solution which gives (II)-(S)(III) pq≠10⇒Dz≠0So, no solution which gives (III)-(S)(IV) pq=10⇒Dx=Dy=Dz=D=0So, infinitely many solutions which give (IV)-(Q),(R),(T)

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Let $p, q, r$ be non-zero real numbers that are respectively the $10^{t h}, 100^{t h} a n d 1000^{t h}$ terms of a harmonic progression. Consider the system of linear equations:
$x + y + z = 1$ , $10 x + 100 y + 1000 z = 0$ , $(q r) x + (p r) y + (p q) z = 0$
Column-I Column-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:

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$(I) - (T); (I I) - (R); (I I I) - (S); (I V) - (T)$

$(I) - (Q); (I I) - (S); (I I I) - (S); (I V) - (R)$

$(I) - (Q); (I I) - (R); (I I I) - (P); (I V) - (R)$

$(I) - (T); (I I) - (S); (I I I) - (P); (I V) - (T)$

answer is B.

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Detailed Solution

(I) $\frac{q}{r} = 10 \Rightarrow D = D_{x} = D_{y} = D_{z} = 0$
So, there will be infinitely many solutions and they will be of the type $(x, y, z) = (\frac{10}{9} + 10 λ, - \frac{1}{9} - 11 λ, λ)$ which gives (I)-(Q),(R),(T)
(II) $\frac{p}{r} \neq 100 \Rightarrow D_{y} \neq 0$
So, no solution which gives (II)-(S)
(III) $\frac{p}{q} \neq 10 \Rightarrow D_{z} \neq 0$
So, no solution which gives (III)-(S)
(IV) $\frac{p}{q} = 10 \Rightarrow D_{x} = D_{y} = D_{z} = D = 0$
So, infinitely many solutions which give (IV)-(Q),(R),(T)

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