SOLUTION: Prove that there exists a linear transformation T : R2 ! R3 such that T(1; 1) = (1; 0; 2) and T(2; 3) = (1;-1; 4): What is T(8; 11)?

Question 1094593: Prove that there exists a linear transformation T : R2 ! R3 such that T(1; 1) = (1; 0; 2) and
T(2; 3) = (1;-1; 4): What is T(8; 11)?
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Let V = {(1; 1), (2; 3)}
let U = {(1 0;2), (1; -1; 4)}
:
check if V is a linearly independent or dependent set of vectors
:
write the vectors of V as a matrix with each vector written as a column
:
| 1 2 |
| 1 3 |
:
transform the matrix to row echelon form
:
r2 - r1 ---> r2
:
| 1 2 |
| 0 1 |
:
r1 - 2r2 ---> r1
:
| 1 0 |
| 0 1 |
:
Rank is 2 implies the vectors are linearly independent, furthermore any set of two linearly independent vectors in R2 spans R2. Hence V is a basis for R2.
:
Since V is a basis, there exists only one linear transformation that maps the vectors of V into the vectors of U.
:
To solve this case, multiply the matrix composed by the vectors of U as columns, by the inverse of the matrix composed by the vectors of V as columns.
:
Note that since the vectors of V form a basis for R2, V is invertible.
:
We invert the following matrix
:
| 1 2 |
| 1 3 |
:
determinant of this matrix = 3 - 2 = 1, and the inverse matrix is
:
| 3 -2 |
| -1 1 |
:
| 1 1 | times | 3 -2 |
| 0 -1 | | -1 1 |
| 2 4 |
:
result is
:
| 2 -1 |
| 1 -1 |
| 2 0 |
:
this is our linear transformation T
:
you can check that T maps the given R2 vectors to the vectors in U, you do this by multiplying our T matrix by the vector from V
:
we follow this procedure for (8, 11) and we get
:
(5, -3, 16)
:

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