Discussion
The value of k for which the system of equations \(x + 2y + kz = 1;\ \ 2x + ky + 8z = 3\) has no solution is
Answer: C
The system has no solution when k = 4.
Two equations in three unknowns either have infinitely many solutions or none. They have none only when the left-hand sides are proportional while the right-hand sides are not. Proportional coefficients need
\[ \frac{2}{1} = \frac{k}{2} = \frac{8}{k} \]
which gives \(k = 4\). Then the first equation doubled reads \(2x + 4y + 8z = 2\), while the second says \(2x + 4y + 8z = 3\), a contradiction, so there is no solution.
For k = 0, 2 or 8 the two coefficient rows are not proportional, the rank is 2, and the system has infinitely many solutions.
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