Business I / Systems of Linear Equations
Practice question · Match the pairs

Match each Rouché–Frobenius situation to its verdict.

  • rank(A)=rank(A∣b)=nrank(A) = rank(A|b) = n
  • rank(A)=rank(A∣b)<nrank(A) = rank(A|b) < n
  • rank(A)<rank(A∣b)rank(A) < rank(A|b)
  • det⁡A≠0\det A \ne 0 (square system)
  • Infinitely many solutions
  • No solution
  • Unique solution guaranteed
  • Unique solution
Hints
  1. Compare the rank of the coefficient matrix with that of the augmented one.
  2. Then compare the common rank with the number of unknowns.
Show the answer
  • rank(A)=rank(A∣b)=nrank(A) = rank(A|b) = n → Unique solution
  • rank(A)=rank(A∣b)<nrank(A) = rank(A|b) < n → Infinitely many solutions
  • rank(A)<rank(A∣b)rank(A) < rank(A|b) → No solution
  • det⁡A≠0\det A \ne 0 (square system) → Unique solution guaranteed
Why

Ranks equal and full: one point. Ranks equal but short: freedom remains. Augmented rank higher: contradiction. And the determinant is the square-system shortcut to case one.

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