Question

Linear Algebra Part B of attachment

Answer 1

Answer to part A by Alchemista:
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)