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: When does a system have a solution.
A system is consistent (has at least one solution) when reducing the equations never gives a false line like $0 = $ a nonzero number. We will reduce and force that bad line to vanish.

Step 2: Subtract to remove $x$.
Subtract the first equation from the second and third. \[ y + 3z = \eta - 1, \qquad 3y + 9z = \eta^2 - 1 \]

Step 3: Make the $y$ terms match.
The third reduced line is just 3 times $(y+3z)$ on the left. So subtract 3 times the first reduced line from it.

Step 4: Look at what is left.
The left side becomes zero, so the right side must also be zero for a solution: \[ (\eta^2 - 1) - 3(\eta - 1) = 0 \]

Step 5: Simplify.
Expand and collect: \[ \eta^2 - 3\eta + 2 = 0 \]

Step 6: Solve the quadratic.
Factor it: \[ (\eta - 1)(\eta - 2) = 0 \implies \eta = 1 \text{ or } \eta = 2 \] \[ \boxed{ \eta = 1 \text{ or } 2 } \]