Matrices and Determinants JEE 2022: HP Consistency Solution

Matrices and Determinants JEE 2022: HP Consistency Match List

JEE Advanced 2022 Mathematics Matrices and Determinants Consistency of system of linear equations with harmonic progression

By Founder, JEEnius - IIT Kanpur Alumni · Aug 21, 2026 · 4 min read

Hard 5 min target

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

$x + y + z = 1$

$10 x + 100 y + 1000 z = 0$

$q r x + p r y + p q z = 0$

Consider the following lists:

| List-I | List-II |
|---|---|
| (I) If $\frac{q}{r} = 10$ , then the system of linear equations has | (P) $x = 0$ , $y = \frac{10}{9}$ , $z = - \frac{1}{9}$ |
| (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 equation 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:

Show answerAnswer

B) (I) $\to$ (Q); (II) $\to$ (S); (III) $\to$ (S); (IV) $\to$ (R)

Explanation

Since $p$ , $q$ , $r$ are terms of a harmonic progression, their reciprocals are terms of an arithmetic progression.

Let

$u = \frac{1}{p}$

$v = \frac{1}{q}$

$w = \frac{1}{r}$

Then $u$ , $v$ , $w$ are the $10^{th}$ , $100^{th}$ , and $1000^{th}$ terms of an arithmetic progression.

For an arithmetic progression, the term is linear in the index. Hence

$v - u = 90 d$

$w - v = 900 d$

$w - v = 10 (v - u)$

Therefore

$w = 11 v - 10 u$

Now divide the third equation by $p q r$ , which is nonzero.

$q r x + p r y + p q z = 0$

After division,

$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$

So the system becomes

$x + y + z = 1$

$10 x + 100 y + 1000 z = 0$

$u x + v y + w z = 0$

Since $u$ , $v$ , $w$ are values of an arithmetic progression at indices $10$ , $100$ , and $1000$ , write the reciprocal AP term as

$b_{n} = A + B n$

Then

$u = A + 10 B$

$v = A + 100 B$

$w = A + 1000 B$

Hence

$u x + v y + w z = (A + 10 B) x + (A + 100 B) y + (A + 1000 B) z$

$u x + v y + w z = A (x + y + z) + B (10 x + 100 y + 1000 z)$

Using the first two equations,

$x + y + z = 1$

$10 x + 100 y + 1000 z = 0$

Therefore

$u x + v y + w z = A$

But the third equation requires

$u x + v y + w z = 0$

So the system is consistent if and only if

$A = 0$

Now find when $A = 0$ . Since

$u = A + 10 B$

$v = A + 100 B$

We have

$v - u = 90 B$

$B = \frac{v - u}{90}$

Also

$A = u - 10 B$

$A = u - \frac{v - u}{9}$

$A = \frac{10 u - v}{9}$

Thus

$A = 0$

is equivalent to

$v = 10 u$

Since

$u = \frac{1}{p}$

$v = \frac{1}{q}$

we get

$\frac{1}{q} = \frac{10}{p}$

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

Also, using $w = 11 v - 10 u$ , if $v = 10 u$ , then

$w = 11 (10 u) - 10 u$

$w = 100 u$

Therefore

$\frac{p}{r} = \frac{w}{u} = 100$

and

$\frac{q}{r} = \frac{w}{v} = 10$

So the following three conditions are equivalent:

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

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

$\frac{q}{r} = 10$

If any one of these holds, then $A = 0$ , the third equation is dependent on the first two equations, and the system has infinitely many solutions. If not, the system has no solution.

For case (I), $\frac{q}{r} = 10$ , so the system is consistent. Also check the solution in (Q):

$x = \frac{10}{9}$

$y = - \frac{1}{9}$

$z = 0$

First equation:

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

$x + y + z = 1$

Second equation:

$10 x + 100 y + 1000 z = 10 \cdot \frac{10}{9} + 100 \cdot (- \frac{1}{9}) + 0$

$10 x + 100 y + 1000 z = \frac{100}{9} - \frac{100}{9}$

$10 x + 100 y + 1000 z = 0$

Since $\frac{q}{r} = 10$ implies $\frac{p}{q} = 10$ , the third equation also holds. Hence (I) matches (Q).

For case (II), $\frac{p}{r} \neq 100$ , so the consistency condition fails. Hence the system has no solution. Thus (II) matches (S).

For case (III), $\frac{p}{q} \neq 10$ , so the consistency condition fails. Hence the system has no solution. Thus (III) matches (S).

For case (IV), $\frac{p}{q} = 10$ , so the consistency condition holds. The third equation becomes dependent on the first two equations. Since two independent linear equations in three variables give infinitely many solutions, (IV) matches (R).

Therefore the correct matching is

$(I) \to (Q)$

$(I I) \to (S)$

$(I I I) \to (S)$

$(I V) \to (R)$

Hence the correct option is B.

What was the JEE Advanced 2022 match-list question on harmonic progression terms and linear system consistency?

p, q, r are the 10th, 100th and 1000th terms of a harmonic progression. The non-homogeneous system is x + y + z = 1, 10x + 100y + 1000z = 0 and qrx + pry + pqz = 0. List-I contains four conditional statements on the ratios q/r, p/r, p/q. List-II offers the specific triple (10/9, -1/9, 0), another triple, infinitely many solutions, no solution, at least one solution. The correct matching from the four options is B.

The three ratio conditions are algebraically identical via the AP relation w = 11v - 10u. One substitution p/q = 10 collapses all four List-I cases into a single insight instead of four separate checks.

How do you solve the matrices and determinants JEE 2022 consistency question using the official method?

Set the reciprocals

u = \frac{1}{p}, v = \frac{1}{q}, w = \frac{1}{r} .

These turn the HP into an AP with v - u = 90d and w - v = 900d. This implies $w = 11 v - 10 u .$

Divide the third equation by nonzero pqr to obtain $u x + v y + w z = 0 .$

Write

u = A + 10 B, v = A + 100 B, w = A + 1000 B .

Then ux + vy + wz reduces exactly to

A (x + y + z) + B (10 x + 100 y + 1000 z) .

Substitute the first two equations to get ux + vy + wz = A. Consistency requires $A = 0$ which simplifies to

\frac{10 u - v}{9} = 0

hence v = 10u or p/q = 10.

When v = 10u the relation forces w = 100u so p/r = 100 and q/r = 10 become equivalent.

For (I) q/r = 10 implies the specific solution x = 10/9, y = -1/9, z = 0 satisfies all three equations. For (II) and (III) the inequality means A ≠ 0 so the third equation contradicts the first two and the system has no solution. For (IV) p/q = 10 makes the third equation linearly dependent giving rank 2 and infinitely many solutions.

The correct matching is (I)→(Q), (II)→(S), (III)→(S), (IV)→(R) which is option B.

Search the past-paper archive for every JEE Advanced paper from 2007 by chapter to practise identical techniques with free worked solutions.

What common mistake leads students to pick the wrong option in this matrices and determinants JEE 2022 question?

Treating the third equation qrx + pry + pqz = 0 as fully independent without dividing by pqr first prevents expressing ux + vy + wz as the exact combination A(first) + B(second).

Students then miss that consistency depends only on whether A = 0. They incorrectly conclude the system is always consistent or always inconsistent instead of conditioning on A = 0.

This produces the wrong assignment of (R) infinitely many solutions to case (II) or (III) where the condition is deliberately violated.

Why are the three ratio conditions algebraically equivalent under the AP relation from the harmonic progression?

From v = 10u substitute into w = 11v - 10u to obtain w = 100u.

This translates back to 1/r = 100/p so p/r = 100 and 1/r = 10/q so q/r = 10.

Thus any one true ratio forces the other two and forces A = 0 simultaneously. This single equivalence replaces four separate checks with one decision.

What are two same-difficulty consistency problems from matrices and determinants for immediate practice?

Practice Question 1: Let a, b, c be the 2nd, 5th and 11th terms of an arithmetic progression. For the 3-variable homogeneous system x + y + z = 0, 2x + 5y + 11z = 0, ax + by + cz = 0 use the same A + Bn parameterization method on the AP indices rather than blind row reduction to find the condition for a non-trivial solution.

Practice Question 2: Examine the system x + y + z = 1, 10x + 20y + 30z = k, (A + B)x + (A + 2B)y + (A + 3B)z = 3 where the third-row coefficients follow A + Bn. The determinant is zero for all A and B yet consistency must still be checked. Apply the identical A + Bn technique to determine for which k the system has solutions.

The parameterization step takes under 90 seconds once the pattern is recognised. After these you will rewrite any third equation as A(first) + B(second) in under 90 seconds, instantly decide infinite versus no solution, and correctly match the entire list without testing random values.

What are the key takeaways for JEE Advanced consistency problems involving sequences?

HP always converts to AP via reciprocals before touching the equations. Any third plane equation can be written as linear combination of the first two if coefficients satisfy the same common difference ratio.

When the constant term on right-hand side produces A = 0 the system drops to rank 2 and yields infinitely many solutions; otherwise it is inconsistent. Never compute full determinant when the first two rows are already multiples of 10, 100, 1000.

Apply this exact rewrite-the-third-equation process to the next similar problem you meet. That single habit decides correct matching in under two minutes.

If a doubt remains on the practice questions, photograph it for a step-by-step solution.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

How do you solve the matrices and determinants jee 2022 consistency question?

Set the reciprocals u=1/p, v=1/q, w=1/r turning the HP into an AP with w=11v-10u. The third equation after dividing by pqr becomes ux+vy+wz=0 which reduces to A(x+y+z) + B(10x+100y+1000z). This equals A after substituting the first two equations. Consistency requires A=0 implying p/q=10 and infinitely many solutions.

Are the ratio conditions q/r, p/r and p/q equivalent in the JEE 2022 HP question?

Yes. From the AP relation w = 11v - 10u, setting v=10u forces w=100u. This makes p/q=10, p/r=100 and q/r=10 all equivalent. Thus one true condition forces A=0 and changes the nature of solutions for the system.

What was the correct matching in the JEE Advanced 2022 matrices and determinants question?

The correct matching is option B with (I) to specific solution (Q), (II) and (III) to no solution (S), and (IV) to infinitely many solutions (R). This follows because only when p/q=10 is the third equation dependent on the first two giving rank 2.

What common mistake do students make in the matrices and determinants jee 2022 question?

Students treat the third equation as independent without dividing by pqr. This prevents them from seeing it as A times the first equation plus B times the second. They miss that consistency depends solely on whether A equals zero or not.

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