Assignment 5

Question 1:
a must be 0 as no squared terms are allowed in a linear transformation. This means
that (a − b + 1) must also be 0 which means b = 1 so (a, b) = (0, 1).
Question 2:

A2020





0
0 1
1 0 0
0 −1 0
a) A2 =  −1 0 0  A3 =  0 1 0  A4 =  0 0 −1 
0 −1
0 0 1
1 0
0

 0
0 −1 0
=  0 0 −1 
1 0
0
b) A is a 120 degrees rotation. This is because after applying the linear
transformation
we arrive at the same matrix. (A4 = A) The rotation is around the
 3 times,

−1

vector v = −1  which can be found by calculating the eigenvector with an eigenvalue
1




−1 −1 0
−1
of 1 so  0 −1 −1  v = 0 which gives v =  −1 .
1
0 −1
1
Question 3:
Preposition 2.3.7 in the coursebook states that the composition of two linear
transformations is also an linear transformation. This means that applying one linear
transformation, f , after the other, g, (g ◦ f ) or the other way around (f ◦ g) both produce
the same linear transformation.
Question 4:
A does not
 have any left
 inverses however it does have
 infinitely many
 right inverses
p
q
p
q
1
0
1
1
0
 0
1 . This is because
1 =
in the form of  0
.
0 1 0
0 1
1 − p −q
1 − p −q
1
Question 5:
If (In + A)−1 = In − A, then that means (In + A)(In − A) = In . Expanding gives
(In + A)(In − A) = In + A − A + A2 = In + On = In . This proves that (In + A) is invertible
and that its inverse is (In − A).
Question 6:
If A = −AT , then AT = −A. For all n × n matrix we have det(M ) = det(M T ). This
means that for matrix A, det(A)= det(AT ) = det(−A) = −1n × det(A) = −det(A)
((−1)n = −1 since n is odd). This means that det(A) = -det(A) so 2*det(A) = 0 which
means that det(A) = 0.
Question 7:
If we substitute In = 1, and A = x. The question becomes (1 + x) · A−1 = 1 so
1
= 1 − x + x2 − x3 + x4 .... Substituting back gives the inverse for In + A to be
A = 1+x
(In + A)−1 = In − A + A2 − A3 + A4 .... (In + A is definitely invertible as an inverse exists)
−1
2