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Let $C$ be an invertible 2x2 matrix such that:

$$C^{-1} \cdot \begin{bmatrix}1 \\ 2\end{bmatrix} = \begin{bmatrix}3 \\ 4\end{bmatrix}$$

$$C^{-2} \cdot \begin{bmatrix}9 \\ 5\end{bmatrix} = \begin{bmatrix}3 \\ 4\end{bmatrix}$$

Find $2\times2$ matrices $A$ and $B$ so that $CA=B$ and solve for $C$.

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  • $\begingroup$ Can you please format your question using LaTex / MathJax? I am having a tough time reading it. Regards $\endgroup$ – Amzoti Mar 11 '13 at 4:27
  • $\begingroup$ I am not familiar with putting in matrices :/ each matrix one column and 2 rowes; hope that helps $\endgroup$ – IndividualThinker Mar 11 '13 at 4:30
  • $\begingroup$ Is $C^{-1}, C^{-2}$ supposed to represent column 1 and column 2? Also, did I capture what you were trying to write? You can find guidance on Latex / MathJax in the FAQ (see link on right of top-of-page). Regards $\endgroup$ – Amzoti Mar 11 '13 at 4:35
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$$\pmatrix{1\cr2\cr}=C\pmatrix{3\cr4\cr}$$

$$\pmatrix{9\cr5\cr}=CC\pmatrix{3\cr4\cr}=C\pmatrix{1\cr2\cr}$$

Now do you see what to use for $A$ and $B$?

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  • $\begingroup$ Got it ! thanks! $\endgroup$ – IndividualThinker Mar 11 '13 at 15:17
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Hint $\rm\ C\,(C^{-1} u = v)\:\Rightarrow\: u\, =\, \color{#C00}{C v}$

And $\rm\ C^2\, (C^{-2}w = v)\:\Rightarrow\: w = C(\color{#C00}{Cv}) = Cu$

Thus $\rm\ C\, (u,v) = (w,u)$

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