The number of values of a for which the system of equations :
`x+y+z=alpha`
`alphax+ 2alphay +3z = -1`
` x+3alphay + 5z =4`
is inconsistent, is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To determine the number of values of \( \alpha \) for which the given system of equations is inconsistent, we need to analyze the system of equations:

1. \( x + y + z = \alpha \)
2. \( \alpha x + 2\alpha y + 3z = -1 \)
3. \( x + 3\alpha y + 5z = 4 \)

We can express this system in matrix form as \( AX = B \), where:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} \alpha \\ -1 \\ 4 \end{bmatrix} \]

For the system to be inconsistent, the determinant of the coefficient matrix \( A \) must be zero, while the augmented matrix must not have a solution.

### Step 1: Calculate the Determinant of Matrix \( A \)

We calculate the determinant \( D \) of matrix \( A \):

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ D = 1 \cdot \begin{vmatrix} 2\alpha & 3 \\ 3\alpha & 5 \end{vmatrix} - 1 \cdot \begin{vmatrix} \alpha & 3 \\ 1 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} \alpha & 2\alpha \\ 1 & 3\alpha \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 2\alpha & 3 \\ 3\alpha & 5 \end{vmatrix} = (2\alpha)(5) - (3)(3\alpha) = 10\alpha - 9\alpha = \alpha \)

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

3. \( \begin{vmatrix} \alpha & 2\alpha \\ 1 & 3\alpha \end{vmatrix} = (\alpha)(3\alpha) - (2\alpha)(1) = 3\alpha^2 - 2\alpha \)

Putting it all together:

\[ D = 1 \cdot \alpha - 1 \cdot (5\alpha - 3) + 1 \cdot (3\alpha^2 - 2\alpha) \]
\[ D = \alpha - 5\alpha + 3 + 3\alpha^2 - 2\alpha \]
\[ D = 3\alpha^2 - 6\alpha + 3 \]

### Step 2: Set the Determinant to Zero

To find the values of \( \alpha \) for which the system is inconsistent, we set the determinant \( D \) to zero:

\[ 3\alpha^2 - 6\alpha + 3 = 0 \]

Dividing the entire equation by 3:

\[ \alpha^2 - 2\alpha + 1 = 0 \]

This can be factored as:

\[ (\alpha - 1)^2 = 0 \]

Thus, we find:

\[ \alpha - 1 = 0 \implies \alpha = 1 \]

### Step 3: Conclusion

Since the determinant is zero for \( \alpha = 1 \), we need to check if the system is inconsistent at this value. We substitute \( \alpha = 1 \) back into the original equations and check if they yield a contradiction.

After substituting \( \alpha = 1 \):

1. \( x + y + z = 1 \)
2. \( x + 2y + 3z = -1 \)
3. \( x + 3y + 5z = 4 \)

We can solve this system, and if it leads to a contradiction, then the system is inconsistent.

### Final Answer

The number of values of \( \alpha \) for which the system of equations is inconsistent is **1**.

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