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Let A and B be invertible matrices such that

\(\mathbf{A} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \text{ and } \mathbf{B} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}.\)
Calculate 

\((\mathbf{A}\mathbf{B})^{-1} \begin{pmatrix} 1 \\ 2 \end{pmatrix}\)

 Mar 1, 2019
 #1
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\((AB)^{-1} = B^{-1}A^{-1}\)

 

\((AB)^{-1}\begin{pmatrix}1\\2\end{pmatrix}=B^{-1}A^{-1}\begin{pmatrix}1\\2\end{pmatrix}\)

 

\(A\begin{pmatrix}2 \\-1\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix} \Rightarrow A^{-1}\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}2\\-1\end{pmatrix}\)

 

\((AB)^{-1}\begin{pmatrix}1\\2\end{pmatrix} = B^{-1}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}1\\3\end{pmatrix}\\ \text{(for reasons identical to that of }A)\)

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 Mar 2, 2019

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