Question

Linear algebra question.... see attachment please

Answer 1

.

Answer 2

Can you see the attachment?

Answer 3

Yes it is possible.

Answer 4

Let's call the matrix on the RHS \(M\), so you have
\[{\bf{AB}}={\bf{M}}\]
If \(|{\bf{A}}|\neq0\), you can find its inverse \({\bf{A}}^{-1}\), which is pretty straightforward since it's 2x2.

Then
\[{\bf{B}}={\bf{A}}^{-1}{\bf{M}}\]
Then you can find the inverse of \(\bf{B}\), provided it's not singular.

Answer 5

Both \(\bf{AB}\) and \(\bf{A}\) have full rank.

Answer 6

^^^ that was my attempt but it did not work out

Answer 7

\[
\det(\mathbf{A})=5-(-3)(-3)=5-9=-4\neq5+9=13
\]

Answer 8

I see that now!!! is my approach correct?

Answer 9

Yes I think.

Answer 10

Ok thank you

Answer 11

\[
\mathbf{A}^{-1}\text{ and }\mathbf{AB}^{-1}\text{ exist.}\\
\mathbf{A}^{-1}\mathbf{AB}=\mathbf{B}\\
\mathbf{B}^{-1}=\left(\mathbf{A}^{-1}\mathbf{AB}\right)^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A}\\
\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)\mathbf{B}=\left(\mathbf{AB}\right)^{-1}\left(\mathbf{A}\mathbf{B}\right)=\mathbf{I}\\
\mathbf{B}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\mathbf{AB}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\left(\mathbf{AB}\left(\mathbf{AB}\right)^{-1}\right)\mathbf{A}=\mathbf{I}\\
\therefore \mathbf{B}^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A}
\]

Answer 12

^^^ I wish I could see it that way.

Answer 13

If I could give you another medal for that answer I would lol