The system of equations
\(x + y + z = 1\)
\(x + 2y + 3z = k\)
\(x + 4y + 7z = k^{2}\)
is consistent. Then the set of all possible real values of k is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\{1,\ 2\}\)
The coefficient matrix is
\( \displaystyle A = \begin{bmatrix}1 & 1 & 1\ 1 & 2 & 3\ 1 & 4 & 7\end{bmatrix} \)
Expanding along the first row:
\( |A| = 1(14 - 12) - 1(7 - 3) + 1(4 - 2) \)
\( = 2 - 4 + 2 = 0 \)
So the system has no unique solution; it is consistent only for the values of k that make the equations agree.
Subtract the first equation from the second, and the second from the third:
\( y + 2z = k - 1 \)
\( 2y + 4z = k^{2} - k \)
The left side of the second line is 2 times the left side of the first, so the right sides must agree in the same way:
\( k^{2} - k = 2(k - 1) \)
\( k^{2} - 3k + 2 = 0 \)
\( (k - 1)(k - 2) = 0 \)
\( k = 1 \text{ or } k = 2 \)
For each of these values the system has solutions: k = 1 is satisfied by x = 1, y = 0, z = 0, and k = 2 by x = 0, y = 1, z = 0. So the set of values is \(\{1,\ 2\}\).
Why the other options are incorrect
Option (2) \(\{-1,\ -2\}\) — Incorrect. These are the roots with the wrong sign. For k = -1 the right sides give \(k^{2} - k = 2\) but \(2(k - 1) = -4\), so the equations clash.
Option (3) \(\{1\}\) — Incorrect. k = 1 works, but so does k = 2 (x = 0, y = 1, z = 0 satisfies all three equations), so this set is incomplete.
Option (4) \(\{0,\ 3\}\) — Incorrect. For k = 0, \(k^{2} - k = 0\) but \(2(k - 1) = -2\); for k = 3, \(k^{2} - k = 6\) but \(2(k - 1) = 4\). The system has no solution for either value.