Q&A: Relationship Between Determinant and Number of Solutions

by Justin Skycak on


In linear algebra, given a square matrix $A$ and a vector $b$ that is the same height as $A,$ how does the determinant of $A$ relate to the number of solutions to the equation $Ax = b?$


Remember that a matrix is invertible if and only if its determinant is nonzero.

If $\det A \neq 0,$ then $A^{-1}$ exists, so the equation $Ax = b$ has a single unique solution $x = A^{-1} b.$

If $\det A = 0,$ then $A^{-1}$ does not exist, there are either no solutions or infinitely many solutions. Why? Intuitively, this is the same kind of behavior that you’d find in the elementary algebra equation $0x = b.$ And more rigorously:

  • Remember that $Ax$ is just a linear combination of the columns of $A.$
  • If $\det A = 0,$ then the columns of $A$ are dependent, which means they don't span the full space of vectors $b,$ and for any vector $b$ that is in the span, there are infinitely many linear combinations of columns of $A$ that produce $b.$
  • So either $b$ is not in not in the span of $A$'s columns (0 solutions), or it is and there are infinitely many ways to combine $A$'s columns to get to $b$ (infinitely many solutions).