Steps to Solve

1. Write the system of equations

The given system of linear equations can be expressed in matrix form as follows:

$$ \begin{pmatrix} 1 & 1 & 0 \ 0 & -1 & 2 \ \alpha & 0 & \beta \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix}

\begin{pmatrix} 3 \ 1 \ \beta \end{pmatrix} $$

2. Analyze the coefficient matrix

To determine the conditions under which the system has at least one solution, we need to check the coefficient matrix ( A ):

$$ A = \begin{pmatrix} 1 & 1 & 0 \ 0 & -1 & 2 \ \alpha & 0 & \beta \end{pmatrix} $$

3. Determine the determinant of the coefficient matrix

For the system to have at least one solution for any values of ( \alpha ) and ( \beta ), the determinant of the coefficient matrix must not be zero (i.e., we cannot have an inconsistent system):

To find the determinant, compute ( \text{det}(A) ):

$$ \text{det}(A) = 1(-1 \cdot \beta - 0 \cdot 2) - 1(0 \cdot \beta - 0 \cdot \alpha) + 0(0 \cdot 2 - \alpha \cdot -1) = -\beta $$

4. Set the determinant condition

For the matrix to be invertible and have at least one solution, we set the determinant condition:

$$ -\beta \neq 0 \implies \beta \neq 0 $$

This shows that for any ( \alpha ), if ( \beta = 0 ), the system may not have a solution.

5. Conclusion of conditions

Therefore, for the system to have at least one solution for every ( \alpha ), we conclude:

• If ( \beta \neq 0 ), there is at least one solution.
• If ( \beta = 0 ), we must check the equations for consistency.