Question

The system of equations \[ x+4y+6z=20,\qquad x+y+z=6,\qquad x+\lambda y+6z=20 \] has

• A. Solution for any value of \(\lambda\)
• B. No solution for \(\lambda=4\)
• C. Unique solution for \(\lambda=4\)
• D. Infinite solutions for \(\lambda\neq4\)

Hint:

Whenever a parameter appears in a system of linear equations, reduce the augmented matrix first. Then examine the special value(s) of the parameter separately to check whether the system is consistent, inconsistent, or has infinitely many solutions.

Answer 1

The Correct Option is A

Concept:

A system of linear equations is consistent if it has at least one solution. We use row operations on the augmented matrix to check consistency for different values of \( \lambda \).

Step 1: Write the augmented matrix.

\[ \left[ \begin{array}{ccc|c} 1 & 4 & 6 & 20 \\ 1 & 1 & 1 & 6 \\ 1 & \lambda & 6 & 20 \end{array} \right] \]

Apply the row operations:

\[ R_2 \rightarrow R_2 - R_1,\qquad R_3 \rightarrow R_3 - R_1. \]

We obtain:

\[ \left[ \begin{array}{ccc|c} 1 & 4 & 6 & 20 \\ 0 & -3 & -5 & -14 \\ 0 & \lambda-4 & 0 & 0 \end{array} \right] \]

Step 2: Consider different values of \( \lambda \).

Case 1: \( \lambda = 4 \)

The third row becomes:

\[ [0 \;\; 0 \;\; 0 \;|\; 0] \]

This does not produce any contradiction, so the system is consistent.

Case 2: \( \lambda \neq 4 \)

The third row gives:

\[ (\lambda-4)y = 0 \]

Hence,

\[ y = 0. \]

Substitute this into the second row:

\[ -5z = -14 \]

Therefore,

\[ z = \frac{14}{5}. \]

Using the equation

\[ x + y + z = 6, \]

we get

\[ x = 6 - \frac{14}{5} = \frac{16}{5}. \]

Thus, the system has a unique solution and is consistent.

Hence, the system is consistent for every value of \( \lambda \).

Answer:

\[ \boxed{\text{(A) Solution for any value of } \lambda} \]