Assignment 5

Question 1:
amust 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:
a) A2=
0 0 1
−100
0−1 0
A3=
1 0 0
0 1 0
0 0 1
A4=
0−1 0
0 0 −1
1 0 0
A2020 =
0−1 0
0 0 −1
1 0 0
b) Ais a 120 degrees rotation. This is because after applying the linear
transformation 3 times, we arrive at the same matrix. (A4=A) The rotation is around the
vector v=
−1
−1
1
which can be found by calculating the eigenvector with an eigenvalue
of 1 so 
−1−1 0
0−1−1
1 0 −1
v= 0 which gives v=
−1
−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:
Adoes not have any left inverses however it does have infinitely many right inverses
in the form of 
p q
0 1
1−p−q
. This is because 1 0 1
0 1 0 
p q
0 1
1−p−q
=1 0
0 1 .
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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×nmatrix we have det(M) = det(MT). This
means that for matrix A, det(A)= det(AT) = det(−A) = −1n×det(A) = −det(A)
((−1)n=−1 since nis 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
A−1=1
1+x= 1 −x+x2−x3+x4.... Substituting back gives the inverse for In+Ato be
(In+A)−1=In−A+A2−A3+A4.... (In+Ais definitely invertible as an inverse exists)
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