Unique Solution Condition for System of Linear Equations

This solution explains how to determine the specific ordered pair (α, β) that guarantees the given system of linear equations has exactly one unique solution. We will analyze the properties related to the coefficient matrix and its determinant.

Understanding Unique Solution Criteria

A system of linear equations, which can be represented in matrix form as $AX = B$, possesses a unique solution if and only if the determinant of the coefficient matrix, denoted as $|A|$, is non-zero. The fundamental condition for a unique solution is:

If the determinant $|A|$ equals zero, the system will typically have either no solutions or infinitely many solutions, but never a unique one.

Matrix Representation and Determinant Calculation

First, let's write down the provided system of equations and identify its coefficient matrix $A$. The given system is:

• $ \alpha x + (\beta+1)y + z = 2 $
• $ 2\alpha x + (\beta+2)y + z = 3 $
• $ \alpha x + \beta y + 2z = 2 $

The coefficient matrix $A$ is constructed using the coefficients of the variables $x, y,$ and $z$ from each equation:

Our next step is to calculate the determinant of this matrix, $|A|$.

Calculating the Determinant of the Coefficient Matrix

We can calculate the determinant $|A|$ using the cofactor expansion method, for instance, along the first row:

Let's compute the determinants of the three $2 \times 2$ submatrices involved:

• Determinant 1: $ \begin{vmatrix} \beta+2 & 1 \\ \beta & 2 \end{vmatrix} = (\beta+2)(2) - (1)(\beta) = 2\beta + 4 - \beta = \beta + 4 $
• Determinant 2: $ \begin{vmatrix} 2\alpha & 1 \\ \alpha & 2 \end{vmatrix} = (2\alpha)(2) - (1)(\alpha) = 4\alpha - \alpha = 3\alpha $
• Determinant 3: $ \begin{vmatrix} 2\alpha & \beta+2 \\ \alpha & \beta \end{vmatrix} = (2\alpha)(\beta) - (\beta+2)(\alpha) = 2\alpha\beta - (\alpha\beta + 2\alpha) = 2\alpha\beta - \alpha\beta - 2\alpha = \alpha\beta - 2\alpha $

Now, substitute these computed determinants back into the formula for $|A|$:

Expand the terms and simplify the expression:

Combine the like terms involving $\alpha\beta$ and $\alpha$:

Factor out the common term $-\alpha$:

Deriving Conditions on $\alpha$ and $\beta$

Recall that for the system to have a unique solution, we must have $|A| \neq 0$. Substituting our result for the determinant, we get:

This inequality holds true if and only if neither of the factors is zero. Therefore, we derive the necessary conditions for $\alpha$ and $\beta$:

• The factor $ \alpha $ must not be zero: $ \alpha \neq 0 $
• The factor $ (\beta + 1) $ must not be zero: $ \beta + 1 \neq 0 $, which implies $ \beta \neq -1 $

In summary, the system of linear equations will have a unique solution only when α is not equal to 0 and β is not equal to -1.

Testing the Ordered Pairs

We now systematically check each of the given ordered pairs $(\alpha, \beta)$ to see which one satisfies both derived conditions ($ \alpha \neq 0 $ and $ \beta \neq -1 $).

From the analysis in the table, we can see that only the ordered pair $(1, 1)$ satisfies both required conditions: $ \alpha \neq 0 $ and $ \beta \neq -1 $. Therefore, the system of linear equations is guaranteed to have a unique solution when $ (\alpha, \beta) = (1, 1) $.