Question

If the following system of linear equations
\[ x-2y+z=5 \] \[ 2x-y+2z=7 \] \[ x+2y+\lambda z=5 \] has a unique solution, then \(\lambda \ne\)

• A. \(1\)
• B. \(-1\)
• C. \(2\)
• D. \(-2\)
• E. \(0\)

Hint:

For uniqueness of solution in linear systems, always check the determinant of the coefficient matrix. If it is non-zero, the system has a unique solution.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A system of linear equations \(AX=B\) has a unique solution if and only if the determinant of the coefficient matrix A is non-zero (i.e., \(|A| \neq 0\)). We need to find the value of \(\lambda\) for which the determinant is zero. The system will have a unique solution for all other values of \(\lambda\).
Step 2: Key Formula or Approach:
The coefficient matrix A for the given system is:
\[ A = \begin{bmatrix} 1 & -2 & 1
2 & -1 & 2
1 & 2 & \lambda \end{bmatrix} \] We need to find \(\lambda\) such that \(|A|=0\). The system has a unique solution for \(|A| \neq 0\).
Step 3: Detailed Explanation:
Calculate the determinant of A:
\[ |A| = 1 \begin{vmatrix} -1 & 2
2 & \lambda \end{vmatrix} - (-2) \begin{vmatrix} 2 & 2
1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & -1
1 & 2 \end{vmatrix} \] \[ |A| = 1((-1)(\lambda) - (2)(2)) + 2((2)(\lambda) - (2)(1)) + 1((2)(2) - (-1)(1)) \] \[ |A| = (-\lambda - 4) + 2(2\lambda - 2) + (4 + 1) \] \[ |A| = -\lambda - 4 + 4\lambda - 4 + 5 \] \[ |A| = 3\lambda - 3 \] For a unique solution, we must have \(|A| \neq 0\).
\[ 3\lambda - 3 \neq 0 \] \[ 3\lambda \neq 3 \] \[ \lambda \neq 1 \] Step 4: Final Answer:
The system has a unique solution for all values of \(\lambda\) except \(\lambda = 1\). Therefore, the condition is \(\lambda \neq 1\). Option (A) is correct.