$x + y + z = 6, \quad x + 2y + 3z = 10, \quad x + 2y + az = \beta$
Step-by-Step Solution
Key Concept: A system has unique, no, or infinitely many solutions depending on whether the determinant is nonzero, or if zero whether the augmented matrix rank equals the coefficient matrix rank.
# Solution: Analysis of the System of Linear Equations
We are given a system of three linear equations:
$$x + y + z = 6 \quad \text{...(1)}$$
$$x + 2y + 3z = 10 \quad \text{...(2)}$$
$$x + 2y + az = \beta \quad \text{...(3)}$$
Our goal is to analyze the consistency and nature of solutions for different values of $a$ and $\beta$.
## Step 1: Simplify the System
<b>Subtract equation (1) from equation (2):</b>
$$(x + 2y + 3z) - (x + y + z) = 10 - 6$$
$$y + 2z = 4 \quad \text{...(4)}$$
<b>Subtract equation (1) from equation (3):</b>
$$(x + 2y + az) - (x + y + z) = \beta - 6$$
$$y + (a-1)z = \beta - 6 \quad \text{...(5)}$$
## Step 2: Analyze the Reduced System
Now we have a simpler system consisting of equations (4) and (5):
$$y + 2z = 4$$
$$y + (a-1)z = \beta - 6$$
<b>Subtract the first from the second:</b>
$$[y + (a-1)z] - [y + 2z] = (\beta - 6) - 4$$
$$(a - 1 - 2)z = \beta - 10$$
$$(a - 3)z = \beta - 10 \quad \text{...(6)}$$
## Step 3: Case Analysis Based on the Value of $a$
### <i>Case A: $a \neq 3$</i>
From equation (6):
$$z = \frac{\beta - 10}{a - 3}$$
This gives a unique value for $z$.
From equation (4):
$$y = 4 - 2z = 4 - 2\cdot\frac{\beta - 10}{a - 3} = \frac{4(a-3) - 2(\beta - 10)}{a-3} = \frac{4a - 12 - 2\beta + 20}{a-3} = \frac{4a - 2\beta + 8}{a-3}$$
From equation (1):
$$x = 6 - y - z = 6 - \frac{4a - 2\beta + 8}{a-3} - \frac{\beta - 10}{a-3}$$
$$x = \frac{6(a-3) - (4a - 2\beta + 8) - (\beta - 10)}{a-3} = \frac{6a - 18 - 4a + 2\beta - 8 - \beta + 10}{a-3}$$
$$x = \frac{2a + \beta - 16}{a-3}$$
<b>Conclusion:</b> When $a \neq 3$, the system has a <i>unique solution</i> for any value of $\beta$.
### <i>Case B: $a = 3$</i>
Equation (6) becomes:
$$0 \cdot z = \beta - 10$$
<b>Sub-case B1: $a = 3$ and $\beta \neq 10$</b>
We get $0 = \beta - 10$ (contradiction), so the system has <i>no solution</i> (inconsistent).
<b>Sub-case B2: $a = 3$ and $\beta = 10$</b>
Equation (6) becomes $0 = 0$ (always true).
The system reduces to:
$$x + y + z = 6$$
$$y + 2z = 4$$
We have 2 independent equations in 3 unknowns. Express in terms of parameter $z = t$:
$$y = 4 - 2t
Correct Answer: [A-q] [B-p, s] [C-p, r] [D-p, r]