Question

For the system of linear equations\(\alpha x+y+z=1, x+\alpha y+z=1, x+y+\alpha z=\beta\) which one of the following statements is NOT correct ?

• A. It has infinitely many solutions if \(\alpha=2 and \beta=-1\)
• B. \(x+y+z=\frac{3}{4} if \alpha=2 and \beta=1\)
• C. It has infinitely many solutions if\(\alpha=1 and \beta=1\)
• D. It has no solution if \(\alpha=-2 and \beta=1\)s

Answer 1

The Correct Option is A

To analyze the given system of linear equations:

\(\alpha x + y + z = 1\)

\(x + \alpha y + z = 1\)

\(x + y + \alpha z = \beta\)

We will consider each condition provided in the options and determine the validity to find out which statement is NOT correct:

1. Given: \(\alpha = 2\) and \(\beta = -1\).

System of equations becomes:

\[ \begin{align*} 2x + y + z &= 1, \\ x + 2y + z &= 1, \\ x + y + 2z &= -1. \end{align*} \]

Calculate the determinant of the coefficient matrix:

\[ \text{Det} = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ \end{vmatrix} = 2(2 \cdot 2 - 1 \cdot 1) - 1(1 \cdot 2 - 1 \cdot 1) + 1(1 \cdot 1 - 2 \cdot 1) = 2(4 - 1) - (2 - 1) + (1 - 2) = 6 - 1 - 1 = 4. \]

Since the determinant is non-zero, the system has a unique solution. Thus, the statement "It has infinitely many solutions" is not correct for this scenario.

1. Given: \( \alpha = 2 \) and \(\beta = 1\).

The system becomes:

\[ \begin{align*} 2x + y + z &= 1, \\ x + 2y + z &= 1, \\ x + y + 2z &= 1. \end{align*} \]

The equations are consistent here, and specific manipulation can be used to calculate that \(x + y + z = \frac{3}{4}\).

1. Given: \(\alpha = 1\) and \(\beta = 1\).

The system of equations becomes:

\[ \begin{align*} x + y + z &= 1, \\ x + y + z &= 1, \\ x + y + z &= 1. \end{align*} \]

Here, all three equations are identical, indicating infinitely many solutions on the line given by \(x + y + z = 1\).

1. Given: \(\alpha = -2\) and \(\beta = 1\).

The system of equations becomes:

\[ \begin{align*} -2x + y + z &= 1, \\ x - 2y + z &= 1, \\ x + y - 2z &= 1. \end{align*} \]

Calculate the determinant of the coefficient matrix:

\[ \text{Det} = \begin{vmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{vmatrix} = -2((-2 \cdot -2) - (1 \cdot 1)) - 1(1 \cdot -2 - 1 \cdot 1) + 1(1 \cdot 1 - (-2 \cdot 1)) = -2(4 - 1) + (2 - 1) + (1 + 2) = -6 + 1 + 3 = -2. \]

Determinant is non-zero, which means the system is consistent, but since all equations do not simultaneously satisfy \(x\), \(y\), and \(z\), it is inconsistent leading to no solution.

Thus, the statement "It has infinitely many solutions if \(\alpha = 2\) and \(\beta = -1\)" is incorrect as the analysis shows the system would have a unique solution. This is the correct answer to the question.