Question

The system of linear equations $\lambda x+2y+2z=5;2\lambda x+3y+5z=8;4x+\lambda y+6z=10$has:

• A. no solution when $\lambda =2$
• B. infinitely many solution when $\lambda =2$
• C. no solution when $\lambda =8$
• D. a unique solution when $\lambda =8$

Answer 1

The correct option is A

no solution when $\lambda =2$

Explanation for the correct option:

Solving the given set of equations:

The given equations are;

$$\begin{gathered} \lambda x+2y+2z=5 \\ 2\lambda x+3y+5z=8 \\ 4x+\lambda y+6z=10 \end{gathered}$$

The determinant form of these equations is

$D=\left[\begin{array}{ccc}\lambda & 2& 2\\ 2\lambda & 3& 5\\ 4& \lambda & 6\end{array}\right]$

Solving the determinant

$$\begin{gathered} D=18\lambda -5{\lambda }^2-24\lambda +40+4{\lambda }^2-24 \\ ⟹𝐷=-{\lambda }^2-6\lambda +16 \\ \mathrm{Now},\mathrm{since}D=0 \\ \Rightarrow {\lambda }^2+6\lambda -16=0 \\ \Rightarrow {\lambda }^2+8\lambda -2\lambda -16=0 \\ \Rightarrow \lambda \left(\lambda +8\right)-2\left(\lambda +8\right)=0 \\ \Rightarrow \left(\lambda +8\right)\left(\lambda -2\right)=0 \\ \Rightarrow \lambda =-8or2 \end{gathered}$$

For $\lambda =2$

$$\begin{gathered} D_1=\left[\begin{array}{ccc}2& 2& 2\\ 4& 3& 5\\ 10& 2& 6\end{array}\right] \\ {D}_1=2\left(18-10\right)-2\left(24-50\right)+2\left(8-30\right) \\ {D}_1=2\left(8\right)-2\left(-26\right)+2\left(-22\right) \\ {D}_1=16+52-44 \\ {D}_1=24\ne 0 \end{gathered}$$

Therefore, the given equations have no solution for $\lambda =2$

Therefore, the correct answer is option (A).

Explanation for the incorrect options:

Option (B):

Since $\lambda =2$ has no solution.

Therefore, option (B) is the incorrect answer.

Option (C):

For $\lambda =8$

$$\begin{gathered} D_1=\left[\begin{array}{ccc}8& 2& 2\\ 16& 3& 5\\ 10& 8& 6\end{array}\right] \\ {D}_1=8\left(18-40\right)-2\left(96-50\right)+2\left(30-128\right) \\ {D}_1=-176-92-196 \\ {D}_1=-463<0 \end{gathered}$$

The equations will have a solution.

Therefore, option (C) is the incorrect option.

Option (D):

Since at $\lambda =8$ $D_1<0$ so it will not have a unique solution.