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 have a solution (consistency), the rank of the coefficient matrix must equal the rank of the augmented matrix.

Step 2: Meaning
This means any row operations that reduce a row of the coefficient matrix to zero must also reduce the corresponding element of the constants column to zero.

Step 3: Analysis
We write the augmented matrix $[A | B]$ and perform row reduction: \[ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 2 & 4 & | & \eta \\ 1 & 4 & 10 & | & \eta^2 \end{pmatrix} \] Performing $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: \[ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & \eta - 1 \\ 0 & 3 & 9 & | & \eta^2 - 1 \end{pmatrix} \] Performing $R_3 \to R_3 - 3R_2$: \[ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & \eta - 1 \\ 0 & 0 & 0 & | & (\eta^2 - 1) - 3(\eta - 1) \end{pmatrix} \] For the system to be consistent, the last entry in the augmented column must be zero: \[ (\eta^2 - 1) - 3(\eta - 1) = 0 \implies \eta^2 - 3\eta + 2 = 0 \] \[ \implies (\eta - 1)(\eta - 2) = 0 \implies \eta = 1 \text{ or } \eta = 2 \]

Step 4: Conclusion
The system of equations is consistent and has a solution only when $\eta = 1$ or $\eta = 2$.

Final Answer: (A)