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I am trying to prove that rref[A|AB] = [I$_{n}$|B], given an invertible matrix A and another matrix B. Note, B does not have to be invertible, but both A and B are n x n matrices.

I understand that rref(A) = I$_{n}$ because matrix A is invertible. However, I am not sure where to begin with proving the result on the right hand side.

As an aside, I tried using two random 2x2 matrices to try to visualize why this might hold:

$A = \begin{bmatrix} 2 & -1 \\ 3 & 0 \end{bmatrix} $ and $B = \begin{bmatrix}4 & 0 \\ 0 & 0 \end{bmatrix} $

But I still don't understand this.

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  • $\begingroup$ Take the matrix $[A|AB]$. Left multiply by $A^{-1}$, and you get $[I|B]$ $\endgroup$
    – user208649
    Mar 31 '15 at 20:31
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As Cuttlefish says in his/her comment, the easiest way when A is invertible is to simply multiply on the left both sides of the "|" in [A|AB] by $A^{-1}$ . This will yield [I$_{n}$|B]. Of course,you first have to establish that A is invertible and that's easy to check by seeing first if $det(A)\neq 0$.

If you want a more concrete way of proving it, you can set up [A|AB] and apply row operations to both sides of the matrix abstract entries $A = \begin{bmatrix} {a_{11},a_{12},......,a_{1n}} \\ {a_{m1},a_{m2},.....,a_{mn}} \end{bmatrix} $ . You then apply row operations to row reduce both sides of the combination matrix. The result of this tedious but effective method would yield [I$_{n}$|B].

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