Question

Consider the system of equations \[ x+y+z=4\mu,\qquad x+2y+2z=10\mu,\qquad x+3y+4\lambda z=\mu^2+15, \] where \(\lambda\) and \(\mu\) are real numbers. Which of the following is correct?

• A. The system is consistent for all values of \(\lambda\) and \(\mu\).
• B. The system has infinite number of solutions if \(\lambda\neq\frac34\) and \(\mu=15\).
• C. The system is inconsistent if \(\lambda=\frac34,\ \mu\neq1\) and \(\mu\neq15\).
• D. The system has unique solutions if \(\lambda=\frac34\) and \(\mu\neq15\).

Hint:

For a system of linear equations: \[ \boxed{|A|\neq0 \Longrightarrow \text{Unique solution}} \] If \[ \boxed{|A|=0,} \] then compare the ranks of the coefficient matrix and augmented matrix to determine whether the system has infinitely many solutions or is inconsistent.

Answer 1

The Correct Option is C

Step 1: Write the coefficient matrix. The coefficient matrix is \[ A= \begin{bmatrix} 1& 1& 1\\ 1& 2& 2\\ 1& 3& 4\lambda \end{bmatrix}. \]

Step 2: Find the determinant of the coefficient matrix. \[ |A| = \begin{vmatrix} 1& 1& 1\\ 1& 2& 2\\ 1& 3& 4\lambda \end{vmatrix} \] Applying the transformation \[ R_2\rightarrow R_2-R_1,\qquad R_3\rightarrow R_3-R_1, \] we get \[ |A| = \begin{vmatrix} 1& 1& 1\\ 0& 1& 1\\ 0& 2& 4\lambda-1 \end{vmatrix} \] \[ =1\left[(4\lambda-1)-2\right] \] \[ =4\lambda-3. \] Hence, \[ |A|=0 \] when \[ \lambda=\frac34. \]

Step 3: Discuss the cases. If \[ \lambda\neq\frac34, \] then \[ |A|\neq0, \] so the system has a unique solution. Now let \[ \lambda=\frac34. \] The equations become \[ x+y+z=4\mu, \] \[ x+2y+2z=10\mu, \] \[ x+3y+3z=\mu^2+15. \] Subtracting the first equation from the second, \[ y+z=6\mu. \] Subtracting the second equation from the third, \[ y+z=\mu^2-10\mu+15. \] For consistency, \[ 6\mu=\mu^2-10\mu+15. \] \[ \mu^2-16\mu+15=0. \] \[ (\mu-1)(\mu-15)=0. \] Thus, \[ \mu=1 \] or \[ \mu=15. \] Therefore, if \[ \lambda=\frac34 \] and \[ \mu\neq1,\qquad \mu\neq15, \] the system is inconsistent. Hence, \[ \boxed{\text{Option (C) is correct}.} \] Thus, \[ \boxed{(C)} \] is the correct answer.