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

Given:

The equations are:

\( 2x - y + 3z = 5, \, 3x + 2y - z = 7, \, 4x + 5y + \alpha z = \beta \).

• Step 1: Calculate the determinant \( \Delta \):

  \( \Delta = \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{vmatrix} \)

  Expanding:

  \( \Delta = 7(\alpha + 5) \)

  Condition for Unique Solution:

  • If \( \Delta = 0 \), then \( \alpha = -5 \).
• Step 2: Calculate the sub-determinants:
  • \( \Delta_1 \):

    \( \Delta_1 = \begin{vmatrix} 5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & -5 \end{vmatrix} = -5(\beta - 9) \)

  • \( \Delta_2 \):

    \( \Delta_2 = \begin{vmatrix} 2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & -5 \end{vmatrix} = 11(\beta - 9) \)

  • \( \Delta_3 \):

    \( \Delta_3 = \begin{vmatrix} 2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta \end{vmatrix} = 7(\beta - 9) \)

• Step 3: Analyze the system:
  • For an inconsistent system: At least one of \( \Delta_1, \Delta_2, \Delta_3 = 0 \).

    When \( \alpha = -5, \, \beta = 8 \): Option A: True.

  • For infinite solutions: \( \Delta_1 = \Delta_2 = \Delta_3 = 0 \), which occurs when \( \beta = 9 \) and \( \alpha = -5 \): Option D: True.
  • For unique solutions: When \( \Delta \neq 0 \), no further constraints apply here.
  • Option B is false.

Final Conclusion:

• \( \alpha = -5, \, \beta = 8 \): Option A: True.
• \( \alpha = -5, \, \beta = 9 \): Option D: True.
• \( \alpha = -5, \, \beta = 8 \): Option C: True.
• Option B: False.

Answer 2

By equation 1 and 3 
\(y+2z=8\)
And \(y=8−2z\)
\(x=−2+z\)
Now putting in equation \(2\) 
\(α(z−2)+β(−2z+8)+7z=3\)
\(⇒(α−2β+7)z=2α−8β+3\)
So equations have unique solution if \(α−2β+7\neq0\)
And equations have no solution if \(α−2β+7=0 \;and \;2α−8β+3\neq0\)
And equations have infinite solution if \(α−2β+7=0\) and \(2α−8β+3=0\)