Question

The system of the equations
\(x + y + z = 6\)
\(x + 2y +   \alpha  z = 5 \)
\(x + 2y + 6z = \)\( \beta\) has

• A. Infinitely many solution for  \(\alpha = 6, \beta= 3\)
• B. Infinitely many solution for \(\alpha = 6, \beta= 5\)
• C. Unique solution for \(\alpha = 6, \beta= 5\)
• D. No solution for \(\alpha = 6, \beta= 5\)

Answer 1

The Correct Option is B

To determine when the system of equations has infinitely many solutions, let's consider the given equations:

1. \(x + y + z = 6\)
2. \(x + 2y + \alpha z = 5\)
3. \(x + 2y + 6z = \beta\)

We need to find the values of \(\alpha\) and \(\beta\) for which the system has infinitely many solutions.

For a system of linear equations to have infinitely many solutions, the equations must be dependent, i.e., one equation must be a linear combination of the others.

First, subtract equation 1 from equation 2:

• \((x + 2y + \alpha z) - (x + y + z) = 5 - 6\)
• \(y + (\alpha - 1) z = -1\)   &;  [Equation 4]

Next, subtract equation 2 from equation 3:

• \((x + 2y + 6z) - (x + 2y + \alpha z) = \beta - 5\)
• \((6 - \alpha)z = \beta - 5\)   &;  [Equation 5]

For the equations to be dependent, equations 4 and 5 must be proportional. Hence, the coefficients of \(z\) in both equations should be equal:

• \(\alpha - 1 = 6 - \alpha\)

Solving the above equation for \(\alpha\):

• \(2\alpha = 7\)
• \(\alpha = \frac{7}{2}\)

Setting \(\alpha = 6\) as given in the options, substitute into Equation 5:

• \((6 - 6)z = \beta - 5\)
• \(0 \cdot z = \beta - 5\)

Thus, for this equation to hold for any value of \(z\), \(\beta - 5\) must be zero, i.e., \(\beta = 5\).

Therefore, the system has infinitely many solutions when \(\alpha = 6\) and \(\beta = 5\), which matches the given correct answer option:

• Infinitely many solutions for \(\alpha = 6, \beta = 5\)

Verification: Setting these values results in the equations being consistent and dependent, confirming multiple solutions exist.