Let \(c_{1}.....c_{n}\) be scalars, not all zero, such that…
2017
Let \(c_{1}.....c_{n}\) be scalars, not all zero, such that \(\sum_{i=1}^{n}c_{i}a_{i}\) where are column vectors in \(a_{i}\).
Consider the set of linear equations
\(Ax = b\)
where \(A=\left [ a_{1}.....a_{n} \right ]\) and \(b=\sum_{i=1}^{n}a_{i}\). The set of equations has
- A.
a unique solution at
\(x=J_{n}\)where\(J_{n}\)denotes a\(n\)-dimensional vector of all 1. - B.
no solution
- C.
infinitely many solutions
- D.
finitely many solutions
Attempted by 84 students.
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Correct answer: C
Solution: Provide why the system has infinitely many solutions.
Given that there exist scalars c1,...,cn not all zero with a linear relation among the column vectors, the columns of A are linearly dependent. This implies A has a nontrivial nullspace (there exists a nonzero vector z with A z = 0).
A particular solution: A times the all-ones vector equals the sum of the columns, so A · (all-ones) = sum_i a_i = b. Thus the all-ones vector is a particular solution.
Nontrivial nullspace: Because the columns are dependent there exists a nonzero vector z with A z = 0, so one can add z to any solution and obtain another solution.
General solution: Every solution has the form x = (all-ones) + z where z is any vector in the nullspace of A. Since the nullspace contains nonzero vectors, there are infinitely many choices for z.
Conclusion: The system is consistent and has infinitely many solutions.
Why a unique solution is not possible: A unique solution would require the nullspace to be trivial (only the zero vector). Here the nullspace is nontrivial, so uniqueness fails.
Why no solution is not possible: b is explicitly the sum of the columns, so b lies in the column space of A; therefore at least one solution exists.
Why 'finitely many solutions' is not a valid outcome here: a linear system cannot have a finite number of solutions greater than one; it must be none, exactly one, or infinitely many.