The system of linear equations in x1, x2, x3 has conditions involving α and β where α, β ∈ R.

Steps to Solve

1. Write the augmented matrix for the system

   We need to form the augmented matrix from the given linear equations: $$ \begin{pmatrix} 1 & 1 & 1 & | & 3 \ 0 & -1 & 1 & | & 1 \ 2 & 3 & \alpha & | & \beta \end{pmatrix} $$

2. Perform row operations to simplify the matrix

   We'll use Gaussian elimination to simplify the matrix. Start by making the leading coefficient of the second row a pivot.

   Subtract 2 times the first row from the third row: $$ R_3 = R_3 - 2R_1 = \begin{pmatrix} 2 - 2(1) & 3 - 2(1) & \alpha - 2(1) & | & \beta - 2(3) \end{pmatrix} = \begin{pmatrix} 0 & 1 & \alpha - 2 & | & \beta - 6 \end{pmatrix} $$

   Now the augmented matrix looks like: $$ \begin{pmatrix} 1 & 1 & 1 & | & 3 \ 0 & -1 & 1 & | & 1 \ 0 & 1 & \alpha - 2 & | & \beta - 6 \end{pmatrix} $$

3. Continue simplification

   Now, add the second row to the third row to eliminate the leading 1 in the second column of the third row: $$ R_3 = R_3 + R_2 = \begin{pmatrix} 0 & 1 - 1 & \alpha - 2 + 1 & | & \beta - 6 + 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & \alpha - 1 & | & \beta - 5 \end{pmatrix} $$

   The augmented matrix now becomes: $$ \begin{pmatrix} 1 & 1 & 1 & | & 3 \ 0 & -1 & 1 & | & 1 \ 0 & 0 & \alpha - 1 & | & \beta - 5 \end{pmatrix} $$

4. Determine the conditions for solutions

   The system has a solution if the last row represents a valid equation. This means we need to look closely at the last row:

   • If $\alpha - 1 \neq 0$, then it is a valid equation.
   • If $\alpha - 1 = 0$, we require $\beta - 5 = 0$ as well, which gives us $\beta = 5$.

5. Conclusion of conditions

   Thus, the conditions are:

   • If $\alpha \neq 1$, then the system has a unique solution.
   • If $\alpha = 1$, then $\beta$ must be equal to 5 for the system to have infinitely many solutions.