query: If $A$ and $B$ are two matrices, and $AB=I$ (of suitable order) is $BA=I$ always true?
positive_passages.docid: 3301
positive_passages.text: If $AB = I$ then $BA = I$
negative_passages.docid: 83475
negative_passages.text: Prove Or Disprove: tr(AB)=tr(A)*tr(B)
negative_passages.docid: 43570
negative_passages.text: Finding two non-zero matrices where $AX = BA$
negative_passages.docid: 55062
negative_passages.text: $2\times 2$ matrix $C=AB-BA$
negative_passages.docid: 34973
negative_passages.text: Is it true that the order of $ab$ is always equal to the order of $ba$?
negative_passages.docid: 147490
negative_passages.text: If matrix $AB=A$, does it mean B must be an identity matrix?
negative_passages.docid: 149553
negative_passages.text: Is it always true that $(A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)$
negative_passages.docid: 15416
negative_passages.text: A weak converse of $AB=BA\implies e^Ae^B=e^Be^A$ from "Topics in Matrix Analysis" for matrices of algebraic numbers.
negative_passages.docid: 139934
negative_passages.text: In $\mathbb{Z}/(n)$, does $(a) = (b)$ imply that $a$ and $b$ are associates?
negative_passages.docid: 9574
negative_passages.text: How to show that $\det(AB) =\det(A) \det(B)$?
negative_passages.docid: 120867
negative_passages.text: How prove this two symmetric matrices $AB=0$
negative_passages.docid: 76200
negative_passages.text: Is the rank of $AB$ always equal to the rank of $BA$?
negative_passages.docid: 27767
negative_passages.text: Are two matrices similar iff they have the same Jordan Canonical form?
negative_passages.docid: 128990
negative_passages.text: Prove that, at least one of the matrices $A+B$ and $A-B$ has to be singular
negative_passages.docid: 49330
negative_passages.text: If for two suitable matrices $B$ and $C$, $AA^TB=AA^TC$ can we conclude that $A^TB=A^TC$?
negative_passages.docid: 62850
negative_passages.text: Linear Algebra: If $A^3 = I$, does $A$ have to be $I$?
negative_passages.docid: 158724
negative_passages.text: Show that $B$ is diagonalizable if If $AB=BA$ and $A$ has distinct real eigenvalues