I was asked to solve an invertible matrix, found the inverse but having trouble putting it into a product of elementary matrices. Can anyone help?A^-1 = [-9/2 7/2][ 4 -3]

I'm guessing you were originally told to find the inverse of[tex]A=\begin{bmatrix}6&7\\8&9\end{bmatrix}[/tex]and you've found the inverse to be[tex]A^{-1}=\begin{bmatrix}-\frac92&\frac72\\4&-3\end{bmatrix}[/tex]I'm also guessing that "product of elementary matrices" includes the decomposition of [tex]A^{-1}[/tex] into lower and upper triangular as well as diagonal matrices.First thing I would do would be eliminate the fractions by multiplying the first row of [tex]A^{-1}[/tex] by 2. In matrix form, this is done by multiplying [tex]A^{-1}[/tex] by[tex]\begin{bmatrix}2&0\\0&1\end{bmatrix}[/tex]which you can interpret as "multiply the first row by 2 and leave the second row alone":[tex]\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}-\frac92&\frac72\\4&-3\end{bmatrix}=\begin{bmatrix}-9&7\\4&-3\end{bmatrix}[/tex]Next, we make the matrix on the right side upper-triangular by eliminating the entry in row 2, column 1. This is done via the product[tex]\begin{bmatrix}1&0\\4&9\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}-\frac92&\frac72\\4&-3\end{bmatrix}=\begin{bmatrix}-9&7\\0&1\end{bmatrix}[/tex]which you can interpret as "leave the first row alone, and replace row 2 by 4(row 1) + 9(row 2)".Lastly, multiply both sides by the inverses of all matrices as needed to isolate [tex]A^{-1}[/tex] on the left side. That is,[tex]\left(\begin{bmatrix}1&0\\4&9\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\right)A^{-1}=\begin{bmatrix}-9&7\\0&1\end{bmatrix}[/tex][tex]\left(\begin{bmatrix}1&0\\4&9\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\right)^{-1}\left(\begin{bmatrix}1&0\\4&9\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\right)A^{-1}=\left(\begin{bmatrix}1&0\\4&9\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\right)^{-1}\begin{bmatrix}-9&7\\0&1\end{bmatrix}[/tex][tex]A^{-1}=\left(\begin{bmatrix}1&0\\4&9\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\right)^{-1}\begin{bmatrix}-9&7\\0&1\end{bmatrix}[/tex]For two invertible matrices [tex]X[/tex] and [tex]Y[/tex], we have [tex](XY)^{-1}=Y^{-1}X^{-1}[/tex], so that[tex]A^{-1}=\begin{bmatrix}2&0\\0&1\end{bmatrix}^{-1}\begin{bmatrix}1&0\\4&9\end{bmatrix}^{-1}\begin{bmatrix}-9&7\\0&1\end{bmatrix}[/tex]Compute the remaining inverses:[tex]\begin{bmatrix}2&0\\0&1\end{bmatrix}^{-1}=\begin{bmatrix}\frac12&0\\0&1\end{bmatrix}[/tex][tex]\begin{bmatrix}1&0\\4&9\end{bmatrix}^{-1}=\begin{bmatrix}1&0\\-\frac49&\frac19\end{bmatrix}[/tex]So we have[tex]\begin{bmatrix}-\frac92&\frac72\\4&-3\end{bmatrix}=\begin{bmatrix}\frac12&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\-\frac49&\frac19\end{bmatrix}\begin{bmatrix}-9&7\\0&1\end{bmatrix}[/tex]