Question

The value of \( \lambda \) for which the following system of equations has unique solution: \[ \lambda x+3y-z=1 \] \[ x+2y+z=2 \] \[ -\lambda x+y+2z=-1 \] are:

• A. \( \lambda\neq \frac{5}{2} \)
• B. \( \lambda\neq \frac{3}{2} \)
• C. \( \lambda\neq \frac{7}{2} \)
• D. \( \lambda\neq -\frac{7}{2} \)

Hint:

For a system of linear equations: \[ AX=B \]
• Unique solution exists when: \[ |A|\neq 0 \]
• No unique solution occurs when: \[ |A|=0 \] In determinant expansion: \[ \begin{vmatrix} a & b & c \end{vmatrix} \] always remember the sign pattern: \[ + \quad - \quad + \] while expanding along a row or column.

Answer 1

The Correct Option is A

Concept: A system of linear equations has a unique solution if and only if the determinant of its coefficient matrix is non-zero. For a system: \[ AX=B \] the condition for unique solution is: \[ |A|\neq 0 \] where \(A\) is the coefficient matrix. If: \[ |A|=0 \] then the system may have either infinitely many solutions or no solution. Thus, in this problem, we first form the coefficient matrix and then compute its determinant.

Step 1: Writing the coefficient matrix The given system is: \[ \lambda x+3y-z=1 \] \[ x+2y+z=2 \] \[ -\lambda x+y+2z=-1 \] Therefore, the coefficient matrix is: \[ A= \begin{bmatrix} \lambda & 3 & -1 \\ 1 & 2 & 1 \\ -\lambda & 1 & 2 \end{bmatrix} \] For a unique solution: \[ |A|\neq 0 \]

Step 2: Evaluating the determinant Compute: \[ |A|= \begin{vmatrix} \lambda & 3 & -1 \\ 1 & 2 & 1 \\ -\lambda & 1 & 2 \end{vmatrix} \] Expand along the first row: \[ |A| = \lambda \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} -3 \begin{vmatrix} 1 & 1 \\ -\lambda & 2 \end{vmatrix} +(-1) \begin{vmatrix} 1 & 2 \\ -\lambda & 1 \end{vmatrix} \] Now evaluate each minor carefully. First Minor \[ \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} =(2)(2)-(1)(1) \] \[ =4-1 \] \[ =3 \] Thus first term becomes: \[ \lambda(3)=3\lambda \] Second Minor \[ \begin{vmatrix} 1 & 1 \\ -\lambda & 2\\ \end{vmatrix} =(1)(2)-(1)(-\lambda) \] \[ =2+\lambda \] Thus second term becomes: \[ -3(2+\lambda) \] \[ =-6-3\lambda \] Third Minor \[ \begin{vmatrix} 1 & 2 \\ -\lambda & 1\\ \end{vmatrix} =(1)(1)-(2)(-\lambda) \] \[ =1+2\lambda \] Thus third term becomes: \[ (-1)(1+2\lambda) \] \[ =-1-2\lambda \]

Step 3: Combining all terms Adding all three parts: \[ |A| = 3\lambda+(-6-3\lambda)+(-1-2\lambda) \] \[ =3\lambda-6-3\lambda-1-2\lambda \] \[ =-7-2\lambda \] \[ =-(2\lambda+7) \]

Step 4: Applying the condition for unique solution For unique solution: \[ |A|\neq 0 \] Therefore, \[ -(2\lambda+7)\neq 0 \] \[ 2\lambda+7\neq 0 \] \[ 2\lambda\neq -7 \] \[ \lambda\neq -\frac72 \] Final Answer: \[ \boxed{\lambda\neq -\frac72} \] Hence, the correct option is: \[ \boxed{(D)} \]