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Step-by-step Solution
Step 1: Write the given system of equations
The system of linear equations is:
(1) $(1 + \alpha)x + \beta y + z = 2$
(2) $\alpha x + (1 + \beta) y + z = 3$
(3) $\alpha x + \beta y + 2z = 2$
Step 2: State the condition for a unique solution
A system of three equations in three unknowns $x, y,$ and $z$ has a unique solution if and only if the determinant of its coefficient matrix is non-zero. That is, if
$\Delta =
\begin{vmatrix}
1 + \alpha & \beta & 1 \\
\alpha & 1 + \beta & 1 \\
\alpha & \beta & 2
\end{vmatrix} \neq 0.$
Step 3: Simplify the determinant condition
Using properties of determinants or by performing appropriate operations, one finds that the determinant condition simplifies to:
$\alpha + \beta \neq -2.$
This means the pair $(\alpha, \beta)$ must satisfy $\alpha + \beta \neq -2$ for the system to have a unique solution.
Step 4: Check the given options
We check each ordered pair $(\alpha, \beta)$ to see if $\alpha + \beta = -2$.
(–3, 1): $-3 + 1 = -2$ (does not satisfy $\alpha + \beta \neq -2$).
(1, –3): $1 + (-3) = -2$ (does not satisfy $\alpha + \beta \neq -2$).
(–4, 2): $-4 + 2 = -2$ (does not satisfy $\alpha + \beta \neq -2$).
(2, 4): $2 + 4 = 6 \neq -2$ (satisfies $\alpha + \beta \neq -2$).
Step 5: Conclude the correct answer
The pair $(2, 4)$ is the only one for which $\alpha + \beta \neq -2$, ensuring the determinant of the coefficient matrix is non-zero. Therefore, the system has a unique solution when $(\alpha, \beta) = (2, 4)$.
