Question

Consider a system of linear equations 3x + y - z = 0, x - $\frac{py}{4}$ + z = 2 and 2x - y + 2z = q where p, q ∈ I and p, q ∈ [1, 10], then identify the correct statement(s).

List-I List-II

(P) Number of ordered pairs (p, q) for which system of equation has unique solution is

(Q) Number of ordered pairs (p, q) for which system of equation has no solution is

(R) Number of ordered pairs (p, q) for which system of equation has infinite solutions is

(S) Number of ordered pairs (p, q) for which system of equation has atleast one solution is

• A. P→4; Q→2; R → 1; S → 5
• B. P→4; Q→3; R → 1; S→4
• C. P→4; Q→3; R→2; S→5
• D. P→4; Q→2; R→2; S→4

Answer 1

Answer: (A)

P→4; Q→2; R→1; S→5

Answer 2

We start with the system

(1) 3x + y – z = 0

(2) x – (p/4)y + z = 2

(3) 2x – y + 2z = q

with p and q integers from 1 to 10.

Step 1: Write the coefficient matrix and compute its determinant.

The coefficient matrix is

A = | 3 1 –1 | | 1 –p/4 1 | | 2 –1 2 |

Its determinant is calculated as

Δ = 3·det|–p/4 1; –1 2| – 1·det|1 1; 2 2| + (–1)·det|1 –p/4; 2 –1|

Compute the 2×2 determinants: det|–p/4 1; –1 2| = (–p/4·2) – (1·(–1)) = –(p/2) + 1 det|1 1; 2 2| = 1·2 – 1·2 = 0 det|1 –p/4; 2 –1| = (1·(–1)) – (–p/4·2) = –1 + (p/2)

Thus, Δ = 3[–(p/2)+1] – 0 – [ (p/2)–1 ] = (–3p/2+3) – (p/2 –1) = –(4p/2) + 4 = 4 – 2p

The determinant Δ = 4 – 2p.

Step 2: Classify the solutions.

• For a unique solution, we require Δ ≠ 0 → 4 – 2p ≠ 0 → p ≠ 2. Since p ∈ {1,3,4,…,10} there are 9 values for p, and for each any q (10 choices) gives 9×10 = 90 ordered pairs.

• For p = 2 the determinant is zero and the system can be dependent or inconsistent. Substitute p = 2 into (2): x – (2/4)y + z = x – (y/2) + z = 2. Now the system becomes: (1) 3x + y – z = 0                       → z = 3x + y (2) x – (y/2) + z = 2 (3) 2x – y + 2z = q

Substitute z = 3x+y into (2): x – (y/2) + (3x+y) = 4x + (y/2) = 2 ⟹ 8x + y = 4 ⟹ y = 4 – 8x Then, z = 3x + (4–8x) = 4 – 5x.

Now substitute into (3): 2x – (4–8x) + 2(4–5x) = 2x – 4 + 8x + 8 – 10x = 4 This gives 4 = q. Thus:

• When p = 2 and q = 4 the system is consistent (the three equations reduce to two independent ones) → infinitely many solutions. (1 ordered pair)
• When p = 2 and q ≠ 4, the system is inconsistent → no solution. There are 9 such ordered pairs.

Finally, the number of ordered pairs having at least one solution = (unique solutions) + (infinite solutions) = 90 + 1 = 91.

Step 3: Match with the given List–II.

List–II is arranged as: 1 9 10 90 91 (Positions: 1st, 2nd, 3rd, 4th, 5th)

Now, assign:

• (P) Unique solution count = 90 → 4th element → “4”
• (Q) No solution count = 9 → 2nd element → “2”
• (R) Infinite solutions count = 1 → 1st element → “1”
• (S) At least one solution count = 91 → 5th element → “5”

Looking at the answer options: (A) P→4; Q→2; R→1; S→5 (B) P→4; Q→3; R→1; S→4 (C) P→4; Q→3; R→2; S→5 (D) P→4; Q→2; R→2; S→4

The correct mapping is given in option (A).

Minimal Explanation:

1. Determinant Δ = 4 – 2p; so for p ≠ 2 (9 values) the system has a unique solution (with any of 10 choices for q → 90 pairs).
2. For p = 2: consistency requires q = 4 giving infinitely many solutions (1 pair) and for q ≠ 4, no solution (9 pairs).
3. At least one solution: 90 + 1 = 91 pairs.
4. In List–II (ordered as 1, 9, 10, 90, 91) the answers for unique, no solution, infinite and at least one solution correspond to positions 4, 2, 1, and 5 respectively.

Thus, option (A) is correct.