Question

If the system of linear equations $x + y + z = 1$, $x + 2y + 4z = \eta$, $x + 4y + 10z = \eta^2$ has a solution, then the value of $\eta$ is:

• A. $1$ or $2$
• B. $1$ or $-2$
• C. $2$ or $-2$
• D. $1$ or $3$

Hint:

Look for linear dependency in the columns or rows: $3 \times (\text{Eq. 2}) - 2 \times (\text{Eq. 1}) = (\text{Eq. 3})$ holds for the LHS coefficients. Hence, the same relationship must hold for the RHS: $3\eta - 2 = \eta^2 \implies \eta^2 - 3\eta + 2 = 0$.

Answer 1

The Correct Option is A

Step 1: Concept

For a system of linear equations to be consistent, the rank of the coefficient matrix must be equal to the rank of the augmented matrix.

Step 2: Meaning

Any row operation that reduces a row of the coefficient matrix to zero must also reduce the corresponding entry in the constants column to zero.

Step 3: Analysis

The augmented matrix is:

\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & \eta \\ 1 & 4 & 10 & \eta^2 \end{array} \right] \]

Apply the row operations: \[ R_2 \rightarrow R_2 - R_1,\qquad R_3 \rightarrow R_3 - R_1 \]

\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & \eta - 1 \\ 0 & 3 & 9 & \eta^2 - 1 \end{array} \right] \]

Now perform: \[ R_3 \rightarrow R_3 - 3R_2 \]

\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & \eta - 1 \\ 0 & 0 & 0 & (\eta^2 - 1) - 3(\eta - 1) \end{array} \right] \]

For consistency, the last entry of the augmented matrix must be zero:

\[ (\eta^2 - 1) - 3(\eta - 1) = 0 \] \[ \eta^2 - 3\eta + 2 = 0 \] \[ (\eta - 1)(\eta - 2) = 0 \] \[ \eta = 1 \quad \text{or} \quad \eta = 2 \]
Step 4: Conclusion

The given system of equations is consistent only when:

\[ \eta = 1 \quad \text{or} \quad \eta = 2 \] 
Final Answer: (A) \( \eta = 1 \text{ or } \eta = 2 \)