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.

  • $\begingroup$ Take the matrix $[A|AB]$. Left multiply by $A^{-1}$, and you get $[I|B]$ $\endgroup$
    – user208649
    Mar 31 '15 at 20:31

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].


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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