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: Write the coefficient matrix.
The system of equations can be written in matrix form as:
\[ \begin{bmatrix} 1 & -2 & 1 2 & -1 & 2 1 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x y z \end{bmatrix} = \begin{bmatrix} 5 7 5 \end{bmatrix} \] Let the coefficient matrix be \(A\).

Step 2: Condition for unique solution.
A system of linear equations has a unique solution if and only if:
\[ |A|\ne 0 \] So we must compute the determinant and ensure it is not zero.

Step 3: Expand the determinant.
\[ |A|= \begin{vmatrix} 1 & -2 & 1 2 & -1 & 2 1 & 2 & \lambda \end{vmatrix} \] Expanding along the first row:
\[ =1\begin{vmatrix}-1 & 2 2 & \lambda \end{vmatrix} -(-2)\begin{vmatrix}2 & 2 1 & \lambda \end{vmatrix} +1\begin{vmatrix}2 & -1 1 & 2 \end{vmatrix} \]

Step 4: Evaluate each minor.
\[ \begin{vmatrix}-1 & 2 2 & \lambda \end{vmatrix}=-\lambda-4 \] \[ \begin{vmatrix}2 & 2 1 & \lambda \end{vmatrix}=2\lambda-2 \] \[ \begin{vmatrix}2 & -1 1 & 2 \end{vmatrix}=4+1=5 \]

Step 5: Substitute back.
\[ |A|=1(-\lambda-4)+2(2\lambda-2)+1(5) \] \[ =-\lambda-4+4\lambda-4+5 \] \[ =3\lambda-3 \]

Step 6: Apply uniqueness condition.
For a unique solution:
\[ |A|\ne 0 \Rightarrow 3\lambda-3\ne 0 \] \[ 3\lambda\ne 3 \Rightarrow \lambda\ne 1 \]

Step 7: Match with the options.
Thus, the value of \(\lambda\) must not be \(1\). Hence, the correct answer is:
\[ \boxed{1} \]