Question

Consider the following system of equations
\(\alpha x + 2y + z = 1\)
\(2\alpha x + 3y + z = 1\)
\(3x + \alpha y + 2z = b\)
For some \(\alpha,\beta∈R\) then which of the following is NOT correct?

• A. It has no solution if α = – 1 and β≠2
• B. It has no solution if α ≠– 1 and β=2
• C. It has no solution if α =3 and for all β≠2
• D. It has no solution if α =– 1 and for all β ∈ R

Answer 1

The Correct Option is C

 To determine which option is NOT correct, we need to analyze the given system of equations conditions for "no solution". The system of linear equations is provided as follows:

\(\begin{align*} \alpha x + 2y + z &= 1 \\ 2\alpha x + 3y + z &= 1 \\ 3x + \alpha y + 2z &= b \end{align*}\)

We must find when this system is inconsistent, i.e., has no solutions. Let's analyze each option individually:

1. Option: It has no solution if \(\alpha = -1\) and \(\beta \neq 2\)
2. Option: It has no solution if \(\alpha \neq -1\) and \(\beta = 2\)
3. Option: It has no solution if \(\alpha = 3\) and for all \(\beta \neq 2\)
4. Option: It has no solution if \(\alpha = -1\) and for all \(\beta \in R\)

From the above analysis, the non-feasible option is the one marked originally as correct: It has no solution if \(\alpha = 3\) and for all \(\beta \neq 2\).