Question

linear algebra mapping

Let F : R^3 → R^3 be a linear mapping . Determine the imaging matrix F of the vector (1, 0 , -1)
depicted at ( 2 , 2, 1 ) , (0 , 1, 2 ) depicted in ( 1 , 2, 1 ) and ( 1, 1, 0) is mapped onto itself.

Answer 1

So we are given the linear transformation of 3 vectors.
(1, 0, -1) ---> (2, 2, 1)

(0, 1, 2) ---> (1, 2, 1)

(1, 1, 0) ---> (1, 1, 0)

Your goal is to find the corresponding transformation matrix

Answer 2

and I do that by checking the conditions?
\[T(a ^{->}+b^{->})= T(a ^{->})+T(b^{->})\]

Answer 3

check this
http://www.calpoly.edu/~brichert/teaching/oldclass/f2002217/handouts/goof.pdf

Answer 4

Nice link, but do I have to do all that to figure out F?

Answer 5

Do you have any other better method ?

Answer 6

Notice that we can write (1, 0, 0) as linear combination of the 3 given vectors :

(1, 0, 0) = 2(1, 0, -1) + 1(0, 1, 2) - 1(1, 1, 0)

Answer 7

Then,

T(1, 0, 0) = 2T(1, 0, -1) + T(0, 1, 2) - T(1, 1, 0)
= 2(2, 2, 1) + (1, 2, 1) - (1, 1, 0)
= (4, 5, 3)

Answer 8

similarly you may try finding T(0, 1, 0) and T(0, 0, 1)

Answer 9

@ParthKohli

okey so we say A is our trans Matrix
\[A \left(\begin{matrix}1 \\ 0 \\ -1\end{matrix}\right) = \left(\begin{matrix}2 \\ 2 \\1 \end{matrix}\right)\]and
\[A \left(\begin{matrix}0 \\ 1 \\ 2\end{matrix}\right) = \left(\begin{matrix}1 \\ 2 \\ 1\end{matrix}\right)\]and also
\[A \left(\begin{matrix}1 \\ 1 \\ 0\end{matrix}\right) = \left(\begin{matrix}1 \\ 1 \\ 0\end{matrix}\right)\]

all together we have

\[A\left[\begin{matrix}1 & 0 &1\\ 0 & 1&1 \\ -1 & 2&0\end{matrix}\right] = \left[\begin{matrix}2 & 1 &1\\ 2 & 2&1 \\ 1 & 1&0\end{matrix}\right]\]

multiply the left with the inverse to solve for A

\[A= \left[\begin{matrix}2 & 1 &1\\ 2 & 2&1 \\ 1 & 1&0\end{matrix}\right]*\left[\begin{matrix}1 & 0 &1\\ 0 & 1&1 \\ 1 & 2&0\end{matrix}\right] ^{-1}\]
solve the inverse and then do the multiplication

\[A=\left[\begin{matrix} 4& -3 &2\\ 5 & -4&3 \\ 3 & -3&2\end{matrix}\right] \]

Answer 10

Nicely done!