To solve the problem of determining the value of \( \frac{\alpha}{\beta} - \frac{\beta}{\alpha} \) for the given system of equations with infinitely many solutions, we must first understand the conditions required for a system of linear equations to have infinitely many solutions.
The given system of equations is:
\[\begin{align*} 2x + 3y - 3z &= 3, \\ x + 2y + \alpha z &= 1, \\ 2x - y + z &= \beta. \end{align*}\]For a system of linear equations to have infinitely many solutions, the rank of the coefficient matrix must equal the rank of the augmented matrix and both must be less than the number of variables.
The coefficient matrix \( A \) and the augmented matrix \( [A|b] \) are:
\[A = \begin{bmatrix} 2 & 3 & -3 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{bmatrix}, \quad [A|b] = \begin{bmatrix} 2 & 3 & -3 & 3 \\ 1 & 2 & \alpha & 1 \\ 2 & -1 & 1 & \beta \end{bmatrix}\]Perform row operations to find the ranks of these matrices.
From row reduction, we aim to make the determinant of the coefficient matrix zero for infinitely many solutions. However, performing above row operations won't directly show when a determinant fails (as we're not reduced to echelon completely by eigenvalues). Rely on expressions forming consistency in all planes (hyperplanes), requiring co-linearity and dependencies.
Calculate the determinant for simplification and dependency:
\[\det(A) = 2\left| \begin{matrix} 2 & \alpha \\ -1 & 1 \end{matrix} \right| - 3\left| \begin{matrix} 1 & \alpha \\ 2 & 1 \end{matrix} \right| - 3\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| \]