1. ## Linear Algebra.Linear Transformation.Help

Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.

2. ## Inverse Transformation

Hello ypatia
Originally Posted by ypatia
Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.
Do you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.

In the question you have here, the transformation can be written in matrix form as

$\displaystyle \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}$

So, to find the inverse transformation, you'll need the inverse of the matrix $\displaystyle \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}$, which is $\displaystyle \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}$

To find its effect on $\displaystyle (x, y, z)$, just do the matrix multiplication:

$\displaystyle \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}$

So there's your answer: $\displaystyle T^{-1}x,y,z) \rightarrow (z,y-z,x-y)$

Hello ypatiaDo you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.

In the question you have here, the transformation can be written in matrix form as

$\displaystyle \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}$

So, to find the inverse transformation, you'll need the inverse of the matrix $\displaystyle \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}$, which is $\displaystyle \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}$

To find its effect on $\displaystyle (x, y, z)$, just do the matrix multiplication:

$\displaystyle \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}$

So there's your answer: $\displaystyle T^{-1}x,y,z) \rightarrow (z,y-z,x-y)$

4. The matrix is $\displaystyle \left( {\begin{array}{*{20}c} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right)$ which is singular.
The matrix is $\displaystyle \left( {\begin{array}{*{20}c} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right)$ which is singular.
Sorry, yes. I mapped $\displaystyle (x,y,z)$ onto $\displaystyle (x+y+z, x+y, x)$, of course.