Download 1. Let A = 1 −1 1 1 0 −1 2 1 1 . a) [2 marks] Find the


1
1. Let A = 1
2

−1 1
0 −1.
1
1
a) [2 marks] Find the determinant of A.
This can be solved by one of two methods, either the checkerboard expansion, or by row reduction. The
answer is det = 5.
b) [2 marks] Find A−1 .

This can be solved by row reduction: (A|I) ∼ (I|A−1 ). The answer is A−1

1
2 1
= 51 −3 −1 2
1 −3 1
 
1
c) [2 marks] Solve the equation Ax = 0.
1
 
 
 
1
1
1
Since Ax = 0 we get A−1 Ax = A−1 0 =⇒ x = A−1 0. Plugging in the value of A−1 found above,
1
1
1

 
 
1
2 1
1
2
and multiplying out, we get x = 51 −3 −1 2 0 = 15 −1.
1 −3 1 1
2
1
2. Let T : R2 → R2 denote the linear transformation
vectors on to the x-axis. Let S : R2 → R2
that projects
x
2x + y
denote the linear transformation that sends
to
.
y
y
a) [2 marks] Find the matrix which represents T .
x
x
1
The projection to the x-axis sends the vector
to
. Since obviously
y
0
0
1 0
matrix is
.
0 0
0 x
x
=
, we get that the
0 y
0
b) [2 marks] Find the matrix which represents S.
2
Since
0
1 x
2x + y
2
=
, we get that the matrix is
1 y
y
0
1
.
1
c) [3 marks] Let U denote the linear transformation given by the composition of the two transformations T
and S, so that U is obtained by first applying T , and then applying S to the result. In other words, U : R2 →
T
S
R2 is given by R2 → R2 → R2 . Then find the matrix that represents U.
We are given that U is defined by first applying
then applying S to the result.
Therefore,
if v is a vector,
T, 1 0
2 1
1 0
v and S(Tv) =
v. Multiplying out the
then Uv = S(Tv). By parts (a) and (b), Tv =
0 0
0 1
0 0
2 0
two matrices gives the matrix for U: it is
.
0 0
Some students multiplied the matrices in the wrong order; in working with linear transformations and
matrices the order of operations matters!
2


1+λ λ+1 0
λ + 2 0 is invertible.
3. [3 marks] Find the values of λ such that A =  2
0
0
1
The easiest way to do this is to use the determinant test for invertible matrices. You can expand the determinant via the checkboard (since the bottom row has mostly zeroes). The determinant turns out to be −λ(λ+1),
which is nonzero if and only if λ 6= 0, −1.
3

1
4. Let A = 1
1
1
1
1

1
1.
1
a) [2 marks] Find a basis for Nul(A) and state the dimension.
You have to solve the
 homogeneous

  system (A|0) and write the answer in vector parametric form. The
−1
−1
general solution is y  1  + z  0 . (There are many different ways to write the solution.) One possible
1
    0
−1
−1
basis is  1  ,  0 , and the dimension is two.
0
1
b) [2 marks] Find a basis for Col(A) and state the dimension.
 
1

The first column is a pivot column, so the basis is 1 and the dimension is one. In this case all the columns
1
are the same, so it’s obvious that any single column is a basis for the column space.
4

  
3
1
c) [5] marks] Show that the set B = {−2 ,  1 } is a basis for NulA.
−1
−2
You have to check 3 things: that the given vectors are actually in the nullspace, that they are linearly independent, and that the number of vectors in B matches the dimension of the space. To check that the vectors
are in the nullspace, just multiply each of the given vectors by A. The answer is zero, so the vectors are in the
null space. They are not multiples of each other, so they are linearly independent. Since there are 2 vectors,
and the dimension is 2, we see that the given vectors form a basis for the null space.
 
5
d) [2 marks] Find the coordinates of the vector  0  with respect to the basis B.
−5


   
3
1
5
You have to solve the equation c1 −2 + c2  1  =  0 . The answer is (c1 , c2 ) = (1, 2).
−1
−2
−5
5
5. Mark the following statements true or false. No explanation is needed.
a) [1 mark] If A, B are invertible n × n matrices, then (A2 BA)−1 = A−1 B−1 (A−1 )2 .
This is true: the inverse of a product is the product of the inverses, in the opposite order (shoes and socks
rule!).
b) [1 mark] Let A denote the matrix corresponding to the linear transformation of R2 which reflects vectors
in the line y = 3x. Then A3 = A.
This is true. Doing a reflection does nothing, so we have A2 = I, and A3 just gives A again.
c) [1 mark] Let A, B denote square n × n matrices. Then A2 − B2 = (A − B)(A + B).
This is false. (A − B)(A + B) = A2 − BA + AB − B2 , but AB 6= BA for general matrices.
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