This should be fairly simple as I know matrix A can be found by $$A = E_1^{-1} E_2^{-1}…E_k^{-1}$$
So it should go that $$A^-$$ can be found by $$E_kE_(k-1)…E_1$$
But for some reason my numbers aren't matching up with the book.
I have a matrix $$\begin{bmatrix}1&0&-1\\0 &6&-1\\0&0&4\end{bmatrix}$$
I find that I can get an Identity Matrix from this matrix by doing (1/6)R2 -> R2, (1/4)R3 -> R3, 1/6R3 + R2 -> R2, R3 + R1 -> R1. From there I can find the inverse of the elementary matrices no problem but for some reason my normal E does not multiply into the inverse. Did I do something wrong in my steps?
It should be noted I need to solve the matrix using elementary matrices.
Best Answer
If $E_1\cdots E_n A = I$, then $A^{-1} = E_1\cdots E_n I$.
So you can transform $A$ into $I$ and simultaneously transform $I$ to $A^{-1}$ using the same transformations:
$$ \left[\begin{array}{rrr|rrr} 1&0&-1 & 1 & 0 & 0\\0 &6&-1& 0 & 1 & 0\\0&0&4& 0 & 0& 1 \end{array}\right] \sim \left[\begin{array}{rrr|rrr} 1&0&-1 & 1 & 0 & 0\\0 &1&-\frac16& 0 & \frac16 & 0\\0&0&1& 0 & 0& \frac14 \end{array}\right]\sim \left[\begin{array}{rrr|rrr} 1&0&0 & 1 & 0 & \frac14\\0 &1&0& 0 & \frac16 & \frac1{24}\\0&0&1& 0 & 0& \frac14 \end{array}\right]$$
so $A^{-1} = \begin{bmatrix} 1 & 0 & \frac14 \\ 0 & \frac16 & \frac1{24} \\ 0 & 0 & \frac14\end{bmatrix}$.
Using elementary matrices with this notation, our transformations are $$L_{3,2}\left(\frac16\right)L_{3,1}(1)D_3\left(\frac14\right)D_2\left(\frac16\right)A = I$$
so \begin{align} A^{-1} &= L_{3,2}\left(\frac16\right)L_{3,1}(1)D_3\left(\frac14\right)D_2\left(\frac16\right)\\ &= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & \frac16 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac14 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & \frac16 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 & \frac14 \\ 0 & \frac16 & \frac1{24} \\ 0 & 0 & \frac14\end{bmatrix} \end{align}