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Question: The value of $\lambda$ such that the system $x - 2y + z = -4, 2x - y + 2z = 2$ and $x + y + \lambda ...

The value of λ\lambda such that the system x−2y+z=−4,2x−y+2z=2x - 2y + z = -4, 2x - y + 2z = 2 and x+y+λz=4x + y + \lambda z = 4 has no solution is

A

3

B

1

C

0

D

-3

Answer

1

Explanation

Solution

We are given the system

{x−2y+z=−4,2x−y+2z=2,x+y+λz=4.\begin{cases} x - 2y + z = -4, \\ 2x - y + 2z = 2, \\ x + y + \lambda z = 4. \end{cases}

The system has no solution when the coefficient matrix is singular and the system is inconsistent. The coefficient matrix is

A=(1−212−1211λ).A = \begin{pmatrix} 1 & -2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & \lambda \end{pmatrix}.

Its determinant is

det⁡(A)=1∣−121λ∣−(−2)∣221λ∣+1∣2−111∣.\det(A) = 1\begin{vmatrix}-1 & 2 \\ 1 & \lambda\end{vmatrix} - (-2)\begin{vmatrix}2 & 2 \\ 1 & \lambda\end{vmatrix} + 1\begin{vmatrix}2 & -1 \\ 1 & 1\end{vmatrix}.

Calculating the minors:

det⁡(−121λ)=−1⋅λ−2⋅1=−λ−2,det⁡(221λ)=2λ−2⋅1=2λ−2,det⁡(2−111)=2⋅1−(−1)⋅1=2+1=3.\begin{aligned} \det\begin{pmatrix}-1 & 2 \\ 1 & \lambda\end{pmatrix} &= -1\cdot\lambda - 2\cdot1 = -\lambda - 2, \\ \det\begin{pmatrix}2 & 2 \\ 1 & \lambda\end{pmatrix} &= 2\lambda - 2\cdot1 = 2\lambda - 2, \\ \det\begin{pmatrix}2 & -1 \\ 1 & 1\end{pmatrix} &= 2\cdot1 - (-1)\cdot1 = 2 + 1 = 3. \end{aligned}

Thus,

det⁡(A)=(−λ−2)+2(2λ−2)+3=−λ−2+4λ−4+3=3λ−3.\det(A) = (-\lambda-2) + 2(2\lambda-2) + 3 = -\lambda - 2 + 4\lambda - 4 + 3 = 3\lambda - 3.

Setting det⁡(A)=0\det(A) = 0 gives:

3λ−3=0⟹λ=1.3\lambda - 3 = 0 \quad\Longrightarrow\quad \lambda = 1.

Now, substituting λ=1\lambda=1 into the system yields:

{x−2y+z=−4,2x−y+2z=2,x+y+z=4.\begin{cases} x - 2y + z = -4, \\ 2x - y + 2z = 2, \\ x + y + z = 4. \end{cases}

Eliminating between equations shows an inconsistency (for instance, subtracting the first from the third gives 3y=83y=8 so y=83y=\frac{8}{3} and then the second equation fails to be satisfied). Hence, the value λ=1\lambda = 1 indeed makes the system inconsistent.