Question

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]

Answer 1

I'm guessing you were originally told to find the inverse of

`A=\begin{bmatrix}6&7\\8&9\end{bmatrix}`

and you've found the inverse to be

`A^(-1)=\begin{bmatrix}-\frac92&\frac72\\4&-3\end{bmatrix}`

I'm also guessing that "product of elementary matrices" includes the decomposition of
`A^(-1)` 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
`A^(-1)` by 2. In matrix form, this is done by multiplying
`A^(-1)` by

`\begin{bmatrix}2&0\\0&1\end{bmatrix}`

which you can interpret as "multiply the first row by 2 and leave the second row alone":

`\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}-\frac92&\frac72\\4&-3\end{bmatrix}=\begin{bmatrix}-9&7\\4&-3\end{bmatrix}`

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

`\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}`

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
`A^(-1)` on the left side. That is,

`\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}`

`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}`

For two invertible matrices
`X` and
`Y`, we have
`(XY)^(-1)=Y^(-1)X^(-1)`, so that

`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}`

Compute the remaining inverses:

`\begin{bmatrix}2&0\\0&1\end{bmatrix}^(-1)=\begin{bmatrix}\frac12&0\\0&1\end{bmatrix}`

`\begin{bmatrix}1&0\\4&9\end{bmatrix}^(-1)=\begin{bmatrix}1&0\\-\frac49&\frac19\end{bmatrix}`

So we have

`\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}`