Assume, that $$A$$ and $$B$$ are matrices. Which of the following statements are true and which are false? Note that such an assertion is proved to be untrue by giving a counterexample, namely concrete matrices.

1. If $$AB$$ = $$B$$ and $$B$$ $$≠$$ $$O$$, then $$A$$ $$=$$ $$I$$

2. If $$AB$$ $$=$$ $$I$$, then $$A$$ is the inverse of $$B$$

This has been driving me nuts... any help on how to proceed this. I just cannot seem to find the correct concrete examples like they're asking.

• (2) is the definition of what it means for $A$ to be the inverse of $B$ or vice versa for $B$ to be the inverse of $A$. As for (1), you should immediately be suspicious. As a hint towards a counterexample... diagonal matrices are particularly easy to multiply. $\begin{bmatrix}a_1&0\\0&a_2\end{bmatrix}\begin{bmatrix}b_1&0\\0&b_2\end{bmatrix}=\begin{bmatrix}a_1b_1&0\\0&a_2b_2\end{bmatrix}$ – JMoravitz Nov 8 at 16:25
• @JMoravitz (2) is not the definition. See this post – Omnomnomnom Nov 8 at 16:32
• Are your matrices in $M_n(K)$, i.e., square matrices? – Dietrich Burde Nov 8 at 16:37
• @Omnomnomnom I don't see where in the page you are referring to, but I clearly remember for square matrices $A$ and $B$ having the inverse defined as such in the textbooks from which I was taught. Granted, if your complaint is how $A$ and $B$ aren't said to be square in this case, then yes, I admit then that $A$ shouldn't have received the name "inverse" of $B$ but instead received the name "left-inverse" of $B$ and that counterexamples would exist, for example $A=[1~0~0~0]$ and $B=[1~0~0~0]^T$, resulting in $AB=$. – JMoravitz Nov 8 at 16:37
• @JMoravitz The conventional definition is that "$B$ is the inverse of $A$ if and only if $BA = AB = I$". I have not yet seen and would be surprised to find a textbook definition that only requires either $BA = I$ or $AB = I$ (even if $A$ and $B$ are said to be square). – Omnomnomnom Nov 8 at 16:40

You said it correctly. One should look for "concrete matrices". Take $$A=\begin{pmatrix} 0 & 0 \cr 0 & 1\end{pmatrix},\; B=\begin{pmatrix} 0 & 0 \cr 1 & 0\end{pmatrix}$$ Then $$AB=B$$ and $$B\neq 0$$, $$A\neq I$$.