The system $Ax = b$ has a UNIQUE solution iff
A$A$ is the identity matrix exactly
B$A$ invertible (square, rank $n$)
C$b = 0$, the homogeneous case
D$\det A = 1$, unit determinant
Answer & Solution
Correct answer: B. $A$ invertible (square, rank $n$)
1. For $Ax = b$ to have a unique solution for every $b$, $A$ must be a square INVERTIBLE matrix.
2. Invertibility = square ($n \times n$) AND $\det A \neq 0$ AND rank = $n$ AND $A$ has trivial null space.
3. The unique solution: $x = A^{-1}b$.
4. If $A$ is not invertible, the system has either NO solutions or INFINITELY MANY.
5. Options A, C, D are too specific or wrong.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §3.6 (Invertible matrix theorem)._
Related questions
Let $A$ be a $3 \times 3$ matrix with eigenvalues 1, 2, 3. Then $\det(A)$ equalsIf $A$ is invertible, the eigenvalues of $A^{-1}$ areThe TRACE of $A = \begin{pmatrix} 2 & 5 \\ 7 & 3 \end{pmatrix}$ isGRAM-SCHMIDT process converts a basis intoTwo vectors $u, v$ are ORTHOGONAL iffAn ORTHOGONAL matrix $Q$ satisfiesFor a SYMMETRIC real matrix $A$, all eigenvalues areCAYLEY-HAMILTON THEOREM states that for any $n \times n$ matrix $A$ with characteristic po