Discussion
Consider the system of linear equations
\(x + 2y + z = 5\)
\(2x + ay + 4z = 12\)
\(2x + 4y + 6z = b\)
The values of \(a\) and \(b\) such that there exists a non-trivial null space and the system admits infinite solutions are
Answer: D
The values are \(a = 4, b = 14\).
A non-trivial null space needs the coefficient matrix to be singular:
\[ \begin{aligned} \det &= 1(6a - 16) - 2(12 - 8) + 1(8 - 2a) \\ &= 4a - 16 = 0 \Rightarrow a = 4 \end{aligned} \]
With a = 4, subtract twice the first equation from the others: the second becomes \(2z = 2\), so z = 1, and the third becomes \(4z = b - 10\). For the system to be consistent, \(4 = b - 10\), so b = 14. Then x + 2y = 4 has infinitely many solutions.
With a = 8 the matrix is non-singular and the solution is unique. With a = 4 and b = 12 the two conditions on z clash (z = 1 and z = 1/2), so there is no solution at all.
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