Question

Let ∶ ℝଶ ⟶ ℝଷ be defined by (ଵ , ଶ ) = (ଵ − ଶ , ଵ , 2ଵ + ଶ ). i) Prove that L is a linear transformation from ℝଶ to ℝଷ . ii) Find basis for the Null space N(L) of L and compute Nullity. iii) Find Rank of L. Further, verify Rank Nullity Theorem.

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.