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Linear Algebra 12 Online
OpenStudy (anonymous):

T:R3 -> R2 is a linear transformation where T(1,1,2)=(2,3) T(1,0,1)=(1,1) and T(1,1,0)=(2,1) Determine T(X,Y,Z) and prove that T(X,Y,Z) is a linear transformation.

OpenStudy (blockcolder):

\[A\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}; A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}; A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}\] If you let \[A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ \end{bmatrix}\] Then you will get two system of equations so that you can solve for the entries of A.

OpenStudy (anonymous):

I don't quite understand. Could you please explain?

OpenStudy (zarkon):

put the following matrix into rref \[\left[\begin{array}{ccccc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]\]

OpenStudy (anonymous):

On it.

OpenStudy (anonymous):

Could you tell me what for, please?

OpenStudy (zarkon):

it is a way of finding the answer...it will give you what T(1,0,0) T(0,1,0) and T(0,0,1) are

OpenStudy (zarkon):

then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)

OpenStudy (anonymous):

Is this correct |dw:1379741449340:dw| ?

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