Solving the Matrix for E, given EB = A.

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A = \begin{bmatrix} 1 & 6 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{bmatrix} And B = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 1 & 1 \\ 1 & 6 & -1 \end{bmatrix}

So the equation EB = A eventually turns into E = AB^-1.

(not sure if I'm supposed to show my steps, but the inverted B matrix I got was: \begin{bmatrix} 7/9 & 1/9 & 1/9 \\ -2/9 & 1/9 & 1/9 \\ -5/9 & 7/9 & -2/9 \end{bmatrix} this might be where my errors came in, but let me know what you guys think or if I should show my steps.

Then, I got this for AB^-1. \begin{bmatrix} 1 & 42/9 & -2/3 \\ 0 & -12/9 & 1/3 \\ 0 & -21/9 & 13/9 \end{bmatrix}

It said this was wrong, so any help is welcome. Again, this is only my second post and the last one was not the best haha. Please let me know how I can improve at all. Thanks!

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You got $B^{-1}$ correct. The error is in calculating $A B^{-1}$, the resulting matrix should be

$ E = \begin{bmatrix} 0 && 0 && 1 \\ 0 && 1 && 0 \\ 1 && 0 && 0 \end{bmatrix} $

But, actually, you don't need to calculate $B^{-1}$ here. If you look closely at $A$ and $B$, you'll see that $B$ is matrix $A$ with rows $(1)$ and $(3)$ flipped, so matrix $E$ must be the matrix above.