Respuesta :
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]
