Key Concept: Conditions for a System of Linear Equations to Have Infinitely Many Solutions

For a system of $n$ linear equations in $n$ variables, represented in matrix form as $AX = B$, to have infinitely many solutions, two primary conditions must be met:

1. Determinant of the Coefficient Matrix is Zero: The determinant of the coefficient matrix, $\text{det}(A)$, must be equal to zero. This implies that the inverse of $A$ does not exist, and Cramer's Rule cannot be directly applied to find unique solutions. Geometrically, this means the planes (for 3 variables) are either parallel or coincident.

2. Consistency (All Associated Determinants are Zero): If $\text{det}(A) = 0$, the system is either inconsistent (no solution) or has infinitely many solutions. To ensure infinitely many solutions, the system must be consistent. This is checked by ensuring that the determinants of the matrices formed by replacing each column of the coefficient matrix with the constant terms (i.e., $\text{det}(A_x)$, $\text{det}(A_y)$, $\text{det}(A_z)$ for a 3x3 system) are all zero. Alternatively, using the concept of rank, $\text{rank}(A) = \text{rank}([A|B]) < n$, where $n$ is the number of variables. This means the rank of the coefficient matrix must be equal to the rank of the augmented matrix, and both must be less than the number of variables. When $\text{det}(A)=0$, $\text{rank}(A) < n$. For infinitely many solutions, $\text{rank}([A|B])$ must also be less than $n$ and equal to $\text{rank}(A)$.

In simpler terms, if, after performing valid row operations or elimination, the system reduces to an identity like $0 \cdot z = 0$ (or $0 = 0$), then there are infinitely many solutions. If it reduces to a contradiction like $0 \cdot z = k$ where $k \neq 0$ (or $0 = k$), then there is no solution.

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Given System of Equations

Let the given system of linear equations be:

$$\begin{align*} 2x + y - z &= 3 \quad &(1) \\ x - y - z &= \alpha \quad &(2) \\ 3x + 3y + \beta z &= 3 \quad &(3) \end{align*}$$

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Method 1: Elimination Technique

We will use