Question

For the system of linear equations 
\(x+y+z=6\) 
\(\alpha x+\beta y+7 z=3\) 
\(x+2 y+3 z=14\). 
which of the following is NOT true ?

• A. The system in inconsistent for α =−5 and β = 8
• B. The system has a unique solution for α =−5 and β = 8
• C. The system has infinitely many solutions for α =−6 and β = 9
• D. The system has infinitely many solutions for α =−5 and β = 9

Hint:

For solving linear systems, always check the determinant or use substitution to determine the type of solutions: unique, none, or infinite.

Answer 1

The Correct Option is C

To determine which statement about the system of linear equations is NOT true, we need to analyze the consistency and solution set of these equations under given conditions. The system of equations is:

• \(x + y + z = 6\)
• \(\alpha x + \beta y + 7z = 3\)
• \(x + 2y + 3z = 14\)

Let's analyze each option:

1. For α = −5 and β = 8, substituting these values into equation 2 gives us:
   • \(−5x + 8y + 7z = 3\)

• Solve equations 1 and 3:
  • \(x + y + z = 6 \quad \text{(Equation 1)}\)
  • \(x + 2y + 3z = 14 \quad \text{(Equation 3)}\)
• \((x + 2y + 3z) - (x + y + z) = 14 - 6\)
• \(y + 2z = 8 \quad \text{(New Equation 4)}\)
• Now, solve Equation 1 and New Equation 4:
  • \(x + (8 - 2z) + z = 6\)
  • \(x + 8 - z = 6\)
  • \(x = z - 2 \quad \text{(Equation 5)}\)
• \(−5(z - 2) + 8(8 - 2z) + 7z = 3\)
• Solving gives \(9z = 41\), which provides a unique solution. Thus, the system is not inconsistent, as it indeed has a unique solution.
• Therefore, the system is inconsistent for α = −5 and β = 8 is a false statement.

1. For α = −6 and β = 9, substituting these values into Equation 2 gives:
   • \(−6x + 9y + 7z = 3\)

Conclusion: The statement that is NOT true is "The system has infinitely many solutions for α = −6 and β = 9", as our analysis clearly supports the singularity or inconsistency under such conditions.