GATE ECE 2011 | Question: 35

Question

The system of equations $$ \begin{aligned} &x+y+z=6 \\ &x+4 y+6 z=20 \\ &x+4 y+\lambda z=\mu \end{aligned} $$ has NO solution for values of $\lambda$ and $\mu$ given by

• A. $\lambda=6, \mu=20$
• B. $\lambda=6, \mu \neq 20$
• C. $\lambda \neq 6, \mu=20$
• D. $\lambda \neq 6, \mu \neq 20$

Answer 1

For no solution $\rho$(A) $\lt$ $\rho$(A¦B) where A¦B is the augmented matrix, A is the coefficient matrix and B is the constant matrix.

then, A = $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 4 & 6 \\ 1 & 4 & \lambda \end{bmatrix}$ and B = $\begin{bmatrix} 6 \\ 20 \\ \mu \end{bmatrix}$

and, A¦B = $\begin{bmatrix} 1 & 1 & 1 & 6 \\ 1 & 4 & 6 & 20 \\ 1 & 4 & \lambda & \mu \end{bmatrix}$

Applying row transformation for converting the matrix into Row Echelon form to get the rank of the matrix:

R₂ → R₂ $-$ R₁

R₃ → R₃ $-$ R₁

A¦B = $\begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 3 & 5 & 14 \\ 0 & 3 & \lambda - 1 & \mu - 6 \end{bmatrix}$

R₃ → R₃ $-$ R₂

A¦B = $\begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 3 & 5 & 14 \\ 0 & 0 & \lambda - 6 & \mu - 20 \end{bmatrix}$

Finally, $\rho$(A) $\lt$ $\rho$(A¦B) when $\lambda$ = 6 ($\rho$(A) = 2) and $\mu$ $\neq$ 20 ($\rho$(A¦B) = 3).