# Proving that matrix $B=C$

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.

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HINTS:

(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 $C=B$.

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