Question

The system of linear equations \( 3x + y - z = 2, x - z = 1 \) and \( 2x + 2y + az = 5 \) has unique solution when:

• A. \( a \neq 3 \)
• B. \( a \neq 4 \)
• C. \( a \neq 5 \)
• D. \( a \neq 2 \)
• E. \( a \neq 1 \)

Hint:

For a system \(AX = B\), a unique solution exists if and only if \(|A| \neq 0\). If \(|A| = 0\), the system may have either no solution or infinitely many solutions.

Answer 1

The Correct Option is D

Concept:
A system of linear equations has a unique solution if and only if the determinant of its coefficient matrix is non-zero (\( \Delta \neq 0 \)). 

Step 1: Construct the coefficient matrix determinant.
The given system is: \[ 3x+y-z=2,\quad x-z=1,\quad 2x+2y+az=5 \] So, the coefficient matrix is: \[ \Delta= \begin{vmatrix} 3 & 1 & -1 \\ 1 & 0 & -1 \\ 2 & 2 & a \end{vmatrix} \] 
Step 2: Expand the determinant carefully.
We expand along the second row (since it contains a zero): \[ \Delta= -1 \begin{vmatrix} 1 & -1 \\ 2 & a \end{vmatrix} + 0\cdot(\cdots) - (-1) \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} \] Now evaluate the minors: \[ \begin{vmatrix} 1 & -1 \\ 2 & a \end{vmatrix} = 1\cdot a - (-1)\cdot 2 = a+2 \] \[ \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} = 3\cdot 2 - 1\cdot 2 = 6-2 = 4 \] Substitute back: \[ \Delta=-1(a+2)+4 \] \[ \Delta=-a-2+4 \] \[ \Delta=2-a \] 
Step 3: Apply the condition for unique solution.
For a unique solution: \[ \Delta \neq 0 \] Therefore, \[ 2-a \neq 0 \] \[ a \neq 2 \] 
Final Answer: \[ \boxed{a \neq 2} \]