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matrix of linear transformation
- Thread starter tinzors
- Start date
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Deleted member 4993
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Hi, I have a problem I need help with:
Find a Linear Transformation L: R3 --> R2 such that
L[1;0;0] = [1;0], L[0;2;1] = [1;1], L[2;1;1] = [0;-2]
Thank you in advance!!
Please define Matrix Linear Transformation for us. Then....
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http://www.freemathhelp.com/forum/th...217#post322217
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Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
Hello, tinzors!
I may have misread the problem.
I am fuzzy on Transformations.
We seek a 3x2 matrix \(\displaystyle L \:=\:\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}\) so that:
. . \(\displaystyle \begin{bmatrix}1&0&0\end{bmatrix}\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix} \:=\:\begin{bmatrix}1&0\end{bmatrix}\) . [1]
. . \(\displaystyle \begin{bmatrix}0&2&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix} \:=\:\begin{bmatrix}1&1\end{bmatrix}\) . [2]
. . \(\displaystyle \begin{bmatrix}2&1&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix} \:=\:\begin{bmatrix}0&\text{-}2\end{bmatrix}\) . [3]
From [1]: .\(\displaystyle [a,\,b] \:=\:[1,\,0]\)
From [2]: .\(\displaystyle [2c+e,\,2d+f] \:=\:[1,\,1]\)
From [3]: .\(\displaystyle [2a+c+e,\,3b+d+f] \:=\:[0,\,\text{-}2] \)
We have a system of equations: .\(\displaystyle \begin{Bmatrix} a \;=\;1 \\ b \;=\;0 \\ 2c+e\;=\;1 \\ 2d+f \;=\;1 \\ 2a+c+e \;=\;0 \\ 2b+d+f \;=\;\text{-}2 \end{Bmatrix}\)
Solve the system: .\(\displaystyle \begin{Bmatrix}a\:=\:1 & b\:=\:0 \\ c\:=\:3 & d\:=\:3 \\ e\:=\:\text{-}5 & f\:=\:\text{-}5 \end{Bmatrix}\)
Therefore: .\(\displaystyle L \:=\:\begin{bmatrix}1&0 \\ 3&3 \\ \text{-}5 & \text{-}5 \end{bmatrix}\)
Am I close?
I may have misread the problem.
I am fuzzy on Transformations.
\(\displaystyle \text{Find a linear transformation }\,L\!:\:R^3 \to R^2\)
\(\displaystyle \text{such that: }\:L[1,0,0] = [1,0],\;\;\; L[0,2,1] = [1,1],\;\; L[2,1,1] = [0,\text{-}2]\)
We seek a 3x2 matrix \(\displaystyle L \:=\:\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}\) so that:
. . \(\displaystyle \begin{bmatrix}1&0&0\end{bmatrix}\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix} \:=\:\begin{bmatrix}1&0\end{bmatrix}\) . [1]
. . \(\displaystyle \begin{bmatrix}0&2&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix} \:=\:\begin{bmatrix}1&1\end{bmatrix}\) . [2]
. . \(\displaystyle \begin{bmatrix}2&1&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix} \:=\:\begin{bmatrix}0&\text{-}2\end{bmatrix}\) . [3]
From [1]: .\(\displaystyle [a,\,b] \:=\:[1,\,0]\)
From [2]: .\(\displaystyle [2c+e,\,2d+f] \:=\:[1,\,1]\)
From [3]: .\(\displaystyle [2a+c+e,\,3b+d+f] \:=\:[0,\,\text{-}2] \)
We have a system of equations: .\(\displaystyle \begin{Bmatrix} a \;=\;1 \\ b \;=\;0 \\ 2c+e\;=\;1 \\ 2d+f \;=\;1 \\ 2a+c+e \;=\;0 \\ 2b+d+f \;=\;\text{-}2 \end{Bmatrix}\)
Solve the system: .\(\displaystyle \begin{Bmatrix}a\:=\:1 & b\:=\:0 \\ c\:=\:3 & d\:=\:3 \\ e\:=\:\text{-}5 & f\:=\:\text{-}5 \end{Bmatrix}\)
Therefore: .\(\displaystyle L \:=\:\begin{bmatrix}1&0 \\ 3&3 \\ \text{-}5 & \text{-}5 \end{bmatrix}\)
Am I close?
HallsofIvy
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So much for "Read before Posting"!