Question

For $a,b \in \mathbb{R}$, consider the system of linear equations: \[ x + y + az = 2, \] \[ 2y + 2z = 1, \] \[ ax + 2z = b \] If the system has infinitely many solutions, then which of the following statements is/are correct?

• A. $a=2, \; b=3$
• B. $a=2, \; b=5$
• C. $a=-1, \; b=-\tfrac{3}{2}$
• D. $a=3, \; b=5$

Hint:

For infinite solutions in linear systems, check $\det(A)=0$ to reduce rank, then ensure consistency by substituting into equations to solve for parameters.

Answer 1

The Correct Option is A, C

Step 1: Write the system in matrix form.
The given equations are: \[ x + y + az = 2, 2y + 2z = 1, ax + 2z = b. \] This can be written as: \[ \begin{bmatrix} 1 & 1 & a \\ 0 & 2 & 2 \\ a & 0 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ b \end{bmatrix}. \]

Step 2: Condition for infinite solutions.
For a system to have infinitely many solutions: \[ \text{rank}(A) = \text{rank}([A|B])<\text{number of variables}. \] Here, number of variables = 3. So, $\det(A)=0$ is necessary.

Step 3: Compute determinant of $A$.
\[ A = \begin{bmatrix} 1 & 1 & a \\ 0 & 2 & 2 \\ a & 0 & 2 \end{bmatrix}. \] \[ \det(A) = 1 \cdot (2 \cdot 2 - 2 \cdot 0) - 1 \cdot (0 \cdot 2 - 2a) + a \cdot (0 \cdot 0 - 2a). \] \[ = 1 \cdot 4 - 1 \cdot (-2a) + a \cdot (-2a). \] \[ = 4 + 2a - 2a^2. \] So, \[ \det(A) = -2a^2 + 2a + 4. \]

Step 4: Solve for $a$.
\[ -2a^2 + 2a + 4 = 0 \;\Rightarrow\; a^2 - a - 2 = 0. \] \[ (a-2)(a+1)=0 \;\Rightarrow\; a=2 \text{or} a=-1. \]



Step 5: Find corresponding $b$ values.
\underline{Case 1: $a=2$}
System becomes: \[ x + y + 2z = 2, 2y + 2z = 1, 2x + 2z = b. \] Divide third equation by 2: \[ x+z = \tfrac{b}{2}. \] From first equation: $x = 2 - y - 2z$. Substituting second eqn $y+z=\tfrac{1}{2} \Rightarrow y=\tfrac{1}{2}-z$. So: \[ x = 2 - \left(\tfrac{1}{2}-z\right) - 2z = \tfrac{3}{2} - z. \] Now $x+z = \tfrac{3}{2}$. But also $x+z=\tfrac{b}{2}$. So: \[ \tfrac{b}{2} = \tfrac{3}{2} \Rightarrow b=3. \] Hence $(a,b)=(2,3)$. \underline{Case 2: $a=-1$}
System becomes: \[ x + y - z = 2, 2y + 2z = 1, -x+2z=b. \] From second: $y+z=\tfrac{1}{2} \Rightarrow y=\tfrac{1}{2}-z$. First eqn: $x+\tfrac{1}{2}-z - z=2 \Rightarrow x= \tfrac{3}{2}+2z$. Third eqn: $-x+2z=b$. Substituting: \[ -(\tfrac{3}{2}+2z)+2z=b \Rightarrow -\tfrac{3}{2}=b. \] So $(a,b)=(-1,-\tfrac{3}{2})$. Final Answer:
\[ \boxed{(a,b)=(2,3)\ \text{and}\ (a,b)=(-1,-\tfrac{3}{2})} \]