Questions: Linear Systems: Notation and Solution Existence

Question 1

A 3×2 matrix A has columns v₁ = [1, 0, 0] and v₂ = [0, 1, 0]. For which right-hand side b does the system Ax = b have a solution?

• A. b = [3, 5, 0] — because b is a linear combination of A's columns
• B. b = [3, 5, 7] — because b has three components matching A's three rows
• C. b = [0, 0, 0] — because only the trivial solution is guaranteed
• D. Any b — a 3×2 system always has at least one solution

Answer: (A)

A solution exists if and only if b lies in the column space of A — the set of all linear combinations of A's columns. Here col(A) = span{[1,0,0], [0,1,0]}, which is the xy-plane in ℝ³. b = [3, 5, 0] = 3v₁ + 5v₂ is in this span, so x = [3, 5] is the unique solution. b = [3, 5, 7] has a nonzero z-component that no combination of A's columns can produce, so there is no solution.

Question 2

A homogeneous system Ax = 0 where A is 4×6 is solved. What can you conclude about the number of solutions?

• A. There are infinitely many solutions, since the null space of a 4×6 matrix is nontrivial
• B. There is exactly one solution: x = 0
• C. There may be zero, one, or infinitely many solutions
• D. There are exactly 2 solutions: x = 0 and one nonzero vector

Answer: (A)

For Ax = 0 (homogeneous system), x = 0 is always a solution — so there is never 'no solution.' A 4×6 matrix has 6 unknowns and at most 4 independent equations, so the null space has dimension at least 6 − 4 = 2. This means there are infinitely many solutions parameterized by a 2-dimensional subspace. The key fact: solution sets are always empty, a single point, or an infinite affine subspace — never exactly 2.

Question 3

If the system Ax = b has more than one solution, it must have infinitely many solutions.

• T. True
• F. False

Answer: True

Any two solutions x₁ and x₂ differ by a vector in the null space of A: x₁ − x₂ ∈ null(A). If the null space contains a nonzero vector v, then x₁ + tv is a solution for every scalar t, generating infinitely many. There is no way to have exactly 2 solutions: once you have 2, you have infinitely many by taking all scalar multiples of the difference in the null space direction.

Question 4

The system Ax = b has no solution if and mainly if A has more rows than columns.

• T. True
• F. False

Answer: False

The existence of solutions depends on whether b lies in the column space of A, not on whether the system is overdetermined. A tall (m > n) system may be consistent if b happens to lie in col(A). Conversely, even a square system (m = n) can have no solution if A is singular and b is not in the column space. Shape alone does not determine consistency.

Question 5

Explain what it means geometrically for the system Ax = b to have no solution, exactly one solution, or infinitely many. What linear-algebraic condition determines which case applies?

Model answer: A solution exists if and only if b is in the column space of A (the span of A's columns). If b is not in col(A), no solution exists. If b is in col(A), the number of solutions depends on the null space of A: if null(A) = {0}, the solution is unique; if null(A) contains nonzero vectors, there are infinitely many solutions parameterized by the null space.

The product Ax is a linear combination of A's columns using entries of x as weights. So 'does Ax = b have a solution?' is exactly 'can b be written as a weighted sum of A's columns?' Once we know a solution x₀ exists, any other solution must differ from x₀ by something in the null space. If null(A) = {0}, x₀ is the only solution. If null(A) is nontrivial, every x₀ + v (for v in null(A)) is also a solution, giving infinitely many. These are the only three possibilities for any linear system.