Question

For 𝛼 , 𝛽 ∈ R, consider the system of linear equations
𝑥 + 𝑦 + 𝑧 = 1 
3𝑥 + 𝑦 + 2𝑧 = 2 
5x + 𝛼𝑦 + 𝛽𝑧 = 3 
Then

• A. for every (𝛼, 𝛽 ) , 𝛼 = 𝛽, the system is consistent
• B. there exists (𝛼, 𝛽 ), satisfying 𝛼 − 2𝛽 + 5 = 0, for which the system has a unique solution
• C. there exists a unique pair (𝛼, 𝛽) for which the system has infinitely many solutions
• D. for every (𝛼, 𝛽), 𝛼 ≠ 𝛽, satisfying 𝛼 − 2𝛽 + 5 = 0, the system has infinitely many solutions

Answer 1

The Correct Option is C

To solve this problem, we need to determine the conditions under which a system of linear equations has a unique solution, no solution, or infinitely many solutions. Let's analyze each step:

The given system of linear equations is:

• \(x + y + z = 1\)
• \(3x + y + 2z = 2\)
• \(5x + \alpha y + \beta z = 3\) 

We want to find conditions on \(\alpha\) and \(\beta\) for different types of solutions. First, let’s consider whether the system is consistent or inconsistent. Systems of equations can have:

• A unique solution if the determinant of the coefficient matrix is non-zero.
• Infinitely many solutions if the system is dependent and consistent.
• No solution if the system is inconsistent.

Let's find the determinant of the coefficient matrix:

1	1	1
3	1	2
5	\(\alpha\)	\(\beta\)

The determinant is calculated as:

\(\Delta = 1\left(1 \cdot \beta - 2 \cdot \alpha\right) - 1\left(3 \cdot \beta - 2 \cdot 5\right) + 1\left(3 \cdot \alpha - 1 \cdot 5\right)\)

Solving this, we get:

\(\Delta = \beta - 2\alpha - 3\beta + 10 + 3\alpha - 5 = -2\beta + \alpha + 5\)

For the system to have infinitely many solutions, this determinant must be zero:

\(\alpha - 2\beta + 5 = 0\)

Only a specific pair \((\alpha, \beta)\) will satisfy this exactly, leading to infinitely many solutions. Evaluating this conditional equation:

• If \(\alpha = 0\) and \(\beta = \frac{5}{2}\), then \(\alpha - 2\beta + 5 = 0\) holds, indicating this unique solution makes the determinant zero.

This pair solves the determinant equation exactly:

\(\alpha = 0, \beta = \frac{5}{2} \Rightarrow 0 - 2 \times \frac{5}{2} + 5 = 0\)

Therefore, the correct answer is:

• There exists a unique pair \((\alpha, \beta)\) for which the system has infinitely many solutions.