Question

Linear Algebra Question on attachment

Answer 1

Ok so this question is really long and im going to need some help understanding it for each part of it ie a-e >_<

Answer 2

Part A)
Map: [a, b, c, d] -> [a + b, 2c - d, a - b - c, a + c + 3d]
T(1, 0, 0, 0) = (1, 0, 1, 1)
T(0, 1, 0, 0) = (1, 0, -1, 0)
T(0, 0, 1, 0) = (0, 2, -1, 1)
T(0, 0, 0, 1) = (0, -1, 0, 3)

\[\left[\begin{matrix}1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 \\ 1 & -1 & -1 & 0 \\ 1 & 0 & 1 & 3 \end{matrix}\right]\]

Is it a bijective map?
Yes, it has full rank, rank=4 (Show this with row reduction)

Answer 3

very nice i understand the map you did there Alchemista although im not entirely sure what you mean by bijective map

Answer 4

Surjective (onto)
Injective (one to one)
Bijective (one to one and onto)

Answer 5

ohh i see very neat thank you very much X3 ill do the second part in a wee sec