Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A,B$ be $n×n$ matrices such that $BA = I_n$, where $I_n$ is the identity matrix.
$(1)$ Suppose that there exists $n×n$ matrix $C$ such that $AC = I_n$. Using properties of the matrix multiplication only (Theorem 2, sec. 2.1, p.113) show that $B = C$.
$(2)$ Show that if $A, B$ satisfy $BA = I_n$, then C satisfying $AC = I_n$ exists. Hint: Consider equation $Ax ̄ = e_i$ for $e_i$ - an elements from the standard basis of $R^n$. Show that this equation has a unique solution for each $e_i$.
Theorem 2

Attempt: $(1)$Ehhm... since $BA=I_n$ then $B$ is the inverse of $A$. Same with $C$. And since a matrix has a unique inverse $B=C$. But, I don't see how to prove it using the properties of multiplication only. Maybe something like this: $$A=B^{-1}I_n=B^{-1}\\A=I_nC^{-1}=C^{-1}\\C^{-1}=B^{-1}\\C=B$$

So, I am not sure about this one because I am using the inverse of a matrix, but I am told to only use the properties of multiplication. Hints please.

$(2)$ Lets say that $BA=I_3$. Then, $\vec e_1=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\; ,\vec e_2=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\; ,\vec e_3=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. Then, $AC=I_3$, where $C=\begin{pmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{pmatrix}$. So, $A\begin{pmatrix} c_{11} \\ c_{21} \\ c_{31} \end{pmatrix}= \vec e_1$. I am trying to use that hint here($A\vec x=e_i$ has a unique solution).How do I show that this system is consistent and has a unique solution. Thanks.

share|cite|improve this question
up vote 5 down vote accepted


(1) You have $AC = I_n$. What happens if you multiply both sides of the equation with $B$ and use $BA=I_n$?

(2) What happens if you multiply both sides of $Ax=e_i$ by $B$ and use $BA=I_n$? Do you get a solution $x$? Is it unique?

Afterwards you have so assemble all $n$ solutions of the equations $Ax=e_i$ into the matrix $C$. (1) then tells you that $C=B$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.