Question

For the system of equations
\( x+\lambda y-z=1 \)
\( x+2y+\lambda z=2 \)
\( x+2y+z=2 \),
which one of the following is NOT correct ?

• A. it has unique solution if \( \lambda \) is not a root of the equation \( t^{2}-3t+2=0 \)
• B. it has unique solution if \( \lambda \) is not a root of the equation \( t^{2}-t-2=0 \)
• C. it has infinitely many solutions if \( \lambda=1 \)
• D. it has no solution if \( \lambda=2 \)

Hint:

Find the determinant of the coefficient matrix, then test the special values \( \lambda=1 \) and \( \lambda=2 \) directly.

Answer 1

The Correct Option is B

Coefficient determinant: \( \Delta=\begin{vmatrix}1&\lambda&-1\\1&2&\lambda\\1&2&1\end{vmatrix}=1(2-2\lambda)-\lambda(1-\lambda)-1(2-2)=\lambda^2-3\lambda+2=(\lambda-1)(\lambda-2) \).
Unique solution iff \( \Delta\ne0 \), i.e. \( \lambda \) is not a root of \( t^2-3t+2=0 \) (roots 1 and 2). So option (1) is correct.
Option (2): the roots of \( t^2-t-2=0 \) are 2 and \( -1 \). The value \( \lambda=1 \) is not a root of this equation, yet \( \Delta=0 \) at \( \lambda=1 \), so the solution is not unique there. Option (2) is NOT correct.
\( \lambda=1 \): the equations become \( x+y-z=1 \), \( x+2y+z=2 \), \( x+2y+z=2 \) - two independent consistent equations in three unknowns, so infinitely many solutions. Option (3) is correct.
\( \lambda=2 \): subtracting the third equation from the second gives \( z=0 \); then \( x+2y=2 \) from the third equation but \( x+2y=1 \) from the first, a contradiction, so no solution. Option (4) is correct.
Answer: (2) it has unique solution if \( \lambda \) is not a root of the equation \( t^{2}-t-2=0 \)