Question

Suppose \( \alpha, \beta, \gamma \in \mathbb{R} \). Consider the following system of linear equations.
\[ x + y + z = \alpha, x + \beta y + z = \gamma, x + y + \alpha z = \beta. \] If this system has at least one solution, then which of the following statements is (are) TRUE?

• A. If \( \alpha = 1 \), then \( y = 1 \)
• B. If \( \beta = 1 \), then \( y = \alpha \)
• C. If \( \beta \neq 1 \), then \( \alpha = 1 \)
• D. If \( y = 1 \), then \( \alpha = 1 \)

Hint:

When solving systems of linear equations, look for patterns and relationships between the variables to find solutions or conditions for solutions.

Answer 1

The Correct Option is A, B

The system is
\[\begin{cases} x+y+z=\alpha \\ x+\beta y+z=\gamma \\ x+y+\alpha z=\beta \end{cases}\]

Subtract the first equation from the second and third:
\[(\beta-1)y=\gamma-\alpha \quad (1)\]
\[(\alpha-1)z=\beta-\alpha \quad (2)\]

(A) If \(\alpha=1\):
From (2): \(0=\beta-1 \Rightarrow \beta=1\).
Then from (1): \((\beta-1)y=0 \Rightarrow \gamma-\alpha=0 \Rightarrow \gamma=1\). - True.

(B) If \(\beta=1\):
From (1): \(0=\gamma-\alpha \Rightarrow \gamma=\alpha\). - True.

(C) If \(\beta\neq1\):
Equation (1) gives \(y=\dfrac{\gamma-\alpha}{\beta-1}\), no need for \(\alpha=1\). - False.

(D) If \(\gamma=1\):
No condition forcing \(\alpha=1\). - False.

Correct options: A and B