chrisplusian
  • chrisplusian
Linear algebra question.... see attachment please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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chrisplusian
  • chrisplusian
1 Attachment
IrishBoy123
  • IrishBoy123
.
chrisplusian
  • chrisplusian
Can you see the attachment?

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thomas5267
  • thomas5267
Yes it is possible.
anonymous
  • anonymous
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.
chrisplusian
  • chrisplusian
1 Attachment
thomas5267
  • thomas5267
Both \(\bf{AB}\) and \(\bf{A}\) have full rank.
chrisplusian
  • chrisplusian
^^^ that was my attempt but it did not work out
thomas5267
  • thomas5267
\[ \det(\mathbf{A})=5-(-3)(-3)=5-9=-4\neq5+9=13 \]
chrisplusian
  • chrisplusian
I see that now!!! is my approach correct?
thomas5267
  • thomas5267
Yes I think.
chrisplusian
  • chrisplusian
Ok thank you
thomas5267
  • thomas5267
\[ \mathbf{A}^{-1}\text{ and }\mathbf{AB}^{-1}\text{ exist.}\\ \mathbf{A}^{-1}\mathbf{AB}=\mathbf{B}\\ \mathbf{B}^{-1}=\left(\mathbf{A}^{-1}\mathbf{AB}\right)^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A}\\ \left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)\mathbf{B}=\left(\mathbf{AB}\right)^{-1}\left(\mathbf{A}\mathbf{B}\right)=\mathbf{I}\\ \mathbf{B}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\mathbf{AB}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\left(\mathbf{AB}\left(\mathbf{AB}\right)^{-1}\right)\mathbf{A}=\mathbf{I}\\ \therefore \mathbf{B}^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A} \]
chrisplusian
  • chrisplusian
^^^ I wish I could see it that way.
chrisplusian
  • chrisplusian
If I could give you another medal for that answer I would lol

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