Problem
Let α, β, and γ be real numbers. Consider the following system of linear equations:
x + 2y + z = 7
x + αz = 11
2x – 3y + βz = γ
Match each entry in List-I to the correct entries in List-II.
| List-I | List-II |
| (P) If β = (7α – 3)/2 and γ = 28, then the system has | (1) a unique solution |
| (Q) If β = (7α – 3)/2 and γ ≠ 28, then the system has | (2) no solution |
| (R) If β ≠ (7α – 3)/2, where α = 1 and γ ≠ 28, then the system has | (3) infinitely many solutions |
| (S) If β ≠ (7α – 3)/2, α = 1 and γ = 28, then the system has | (4) x = 11, y = -2, and z = 0 as a solution |
| (5) x = -15, y = 4, and z = 0 as a solution |
Solution
We’ll start with applying the Cramer’s rule.
Using standard notation, Δ = \( \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \\ \end{vmatrix} \) or 7α – 2β – 3.
Let’s calculate Δ1, Δ2, and Δ3 as well.
Δ1 = \( \begin{vmatrix} 7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta \\ \end{vmatrix} \) or 21α – 22β + 2αγ – 33.
Δ2 = \( \begin{vmatrix} 1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta \\ \end{vmatrix} \) or 4β – αγ + γ + 14α – 22.
Δ3 = \( \begin{vmatrix} 1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma \\ \end{vmatrix} \) or -2γ + 56.
We’re ready. Let’s look at each option in the first column, one by one.
(P) We have β = (7α – 3)/2 and γ = 28. This means Δ = Δ1 = Δ2 = Δ3 = 0, which means the system either has infinitely many solutions or no solution. A little more work is needed here.
Let’s substitute the above values in the equations. We’ll get:
x + 2y + z = 7
x + αz = 11
4x – 6y + (7α – 3)z = 56
Now, it’s clear that the first two planes aren’t parallel, and will have a line of intersection.
If we take a closer look at the third equation, it can be written as 7(x + αz – 11) – 3(x + 2y + z – 7) = 0.
That is, a linear combination of the first two equations, which means the plane represented by the third equation passes through the line of intersection of the first two.
This means that the three planes have a common line of intersection. In other words, the three equations have infinitely many solutions. So, (P) matches with (3).
(Q) Here, β = (7α – 3)/2 and γ ≠ 28. This means Δ = 0 but Δ3 ≠ 0, which means the system has no solution. So, (P) matches with (2).
For both (R) and (S), β ≠ (7α – 3)/2. This means Δ ≠ 0, and the system has a unique solution.
This solution is given by x = Δ1/Δ, y = Δ2/Δ, and z = Δ3/Δ.
Now, if α = 1, then Δ = 4 – 2β, Δ1 = -22β + 2γ – 12, Δ2 = 4β – 8, and Δ3 = -2γ + 56.
Also, since β ≠ (7α – 3)/2, we’ll get β ≠ 2.
This means x = (11β – γ + 6)/(β – 2), y = -2, and z = (γ – 28)/(β – 2).
In (S), we have γ = 28. Substituting this in the above expressions, we’ll get x = 11 and z = 0. This means that (S) matches with (4).
In (R), we have γ ≠ 28, which means that z ≠ 0. Therefore, (R) matches with (1).
Summarizing the answers: (P) → (3), (Q) → (2), (R) → (1), (S) → (4)
Note that (S) also matches with (1), but the problem asked to match each option in the left column with only one option in the right one. Plus, we had to choose from one of the given options, and the above was one of them.
Comments
A direct problem on the application of Cramer’s rule. A bit tedious though, given it was just worth 3 marks.
Some work could have been avoided using the options, since two of them had (P) → (3), (Q) → (2), and (S) → (4). So after verifying (P) → (3), (Q) → (2), we only had to check for (R) and skipped (S).
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