Question

Let $S$ be the set of real values of $k$ for which the system of equations \[ kx + y + z = k \] \[ x + ky + z = k \] \[ x + y + kz = k \] has no solution. Then $|S|$ is

• A. 1
• B. 2
• C. 0
• D. $\infty$

Hint:

For symmetric systems, $k = 1$ often represents a case of identical planes (infinitely many solutions), while the other root (here, $k = -2$) represents parallel but distinct conditions (no solution).

Answer 1

The Correct Option is A

Step 1: Decide when the system can fail to have a unique answer.
A unique solution exists when the coefficient determinant is nonzero. So trouble (either no solution or infinitely many) can happen only where the determinant is zero.

Step 2: Compute the determinant.
The coefficient matrix has $k$ on the diagonal and $1$ elsewhere. Its determinant is $(k - 1)^2(k + 2)$.

Step 3: Find the special values of $k$.
Setting $(k - 1)^2(k + 2) = 0$ gives $k = 1$ or $k = -2$. These are the only candidates for no solution.

Step 4: Test $k = 1$.
All three equations become $x + y + z = 1$, the very same equation three times. That is consistent with infinitely many solutions, not no solution. So $k = 1$ is ruled out.

Step 5: Test $k = -2$.
Adding all three equations gives $0 \cdot (x + y + z) = -6$ after simplification, since the left sides cancel to zero but the right sides add to $3(-2) = -6$. A statement like $0 = -6$ is impossible, so there is no solution here.

Step 6: Count the valid values.
Only $k = -2$ gives no solution, so the set $S$ has exactly one element, $|S| = 1$.
\[ \boxed{1} \]