Math 2270
Linear Algebra
Test 1
You will have 50 minutes to complete this test. You may not use a calculator or any other electronic device on
this test. Each problem is wort 10 points, for a total of 100. If there are two parts, each will be graded out of
five points. Good Luck!
1. Find all solutions of the linear system
x + 2y + 3z = 0
4x + 5y + 6z = 0
7x + 8y + 10z = 0
.
We can do this by forming the augmented matrix for the system and row reduce as follows:

 
 
 

..
..
..
..
1
2
3
.
0
1
2
3
.
0
1
0
−1
.
0
1
0
0
.
0

 
 
 


 
 
 

.
.
.
.
 4 5 6 .. 0  ∼  0 −3 −6 .. 0  ∼  0 1 2 .. 0  ∼  0 1 0 .. 0  .

 
 
 

.
.
.
.
7 8 10 .. 0
0 −6 −11 .. 0
0 0 1 .. 0
0 0 1 .. 0
The solution to the system is x1 = x2 = x3 = 0.
2. Row reduce the corresponding agumented matrix to find all solutions of the linear system
3x + 4y − z = 8 6x + 8y − 2z = 3 .
We form the augmented matrix for the system and row reduce it:

 
..
1
4
3
4
−1
.
8

 ∼  1 3 −3
.
0 0
0
6 8 −2 .. 3

..
8
.
3
.
..
. −13
This system does not have any solutions, since the last equation implies that 0 = −13.
3. (a) Let A be an n × n matrix, and let ~b and ~c be two vectors in Rn . We are told that the system A~x = ~b has
a unique solution. What can you say about the number of solutions of the system A~x = ~c?
(b) Now suppose that A is n × m with n > m. We are told that the system A~x = ~0 has a unique solution. If
~b 6= ~0, then does the system A~x = ~b have to have a unique solution?
(a) If A~x = ~b has a unique solution, then rref(A) must have a leading one in each column. Therefore, as A has
the same number of rows and columns, it must be the case that there are no rows of zeros in rref(A), and also no
free variables. Therefore A~x = ~c also has a unique solution.
(b) If A~x = ~0 has a unique solution, then there must be a leading one in each column of rref(A). Therefore, if
A~x = ~b has a solution, it must be unique, but it may not have a solution since there will be at least one row of
zeros in rref(A).
4. Describe all linear transformations from R2 to R. What do their graphs look like?
Suppose that T : R2 → R is a linear transformation. Then, as was proved in class, we can write T (~x) = A~x,
where A is a 1 × 2 matrix. Suppose A = [a1 , a2 ]. Then T (~x) = x1 a1 + x2 a2 . The graph of this function is a plane
through the origin.
1
5. Consider the linear transformation that rotates vectors in R2 by θ radians in the counterclockwise direction.
The matrix corresponding to this transformation is given by
cos θ − sin θ
A=
.
sin θ
cos θ
Show that the matrix A preserves length, i.e. show that k~xk = kA~xk for any ~x ∈ R2 .
We have that
A~x =
x1 cos θ − x2 sin θ
x1 sin θ + x2 cos θ
,
now we compute the length of A~x,
kA~xk2 = (x1 cos θ − x2 sin θ)2 + (x1 sin θ + x2 cos θ)2
= x21 cos2 θ − x1 x2 sin θ cos θ + x22 sin2 θ + x21 sin2 θ + x1 x2 sin θ cos θ + x22 cos2 θ
= (x21 + x22 ) cos2 θ + (x21 + x22 ) sin θ = x21 + x22 = k~xk2 .
6. Compute the inverse of the matrix


1
3 .
6
1 1
A= 1 2
1 3
We form the augmented matrix

 1

 1

1

 1

∼ 0

0
.
[A..I] and row reduce:
 
..
1 1 . 1 0 0   1 1
 
.
∼
2 3 .. 0 1 0 
  0 1
.
3 6 .. 0 0 1
0 2
.
0 −1 .. 2
.
1 2 .. −1
.
0 1 .. 1
1
2
5
 
−1 0   1 0
 
∼
1 0 
  0 1
0 0
−2 1
We have that
A−1


..
. 1 0 0 

..
. −1 1 0 

..
. −1 0 1

.
0 .. 3 −3 1 

.
.
0 .. −3 5 −2 

.
1 .. 1 −2 1

−3 1
5 −2  .
−2 1
3
=  −3
1
7. Compute the matrix product
1 −2
−2 5
−5
11
We have that
1 −2 −5
−2 5
11


8 −1
 1 2 .
1 −1


8 −1
 1 2 = 1
0
1 −1
8. Find vectors that span ker(A), where

1
A= 0
0
2
1
0
2

3 4
2 3 .
0 1
0
1
.
To find these vectors, we

1 2
 0 1
0 0
Let x3 = t. Then
so
solve the homogeneous system A~x = ~0:
 
 
3 4
1 0 −1 −2
1 0
2 3 ∼ 0 1 2
3 ∼ 0 1
0 1
0 0 0
1
0 0

−1 0
2 0  ⇒
0 1


 
1
x1
t
 −2
 x2   −2t 



 
 x3  =  t  = t  1
0
x4
0

x1 = x3
x2 = −2x3 .
x4 = 0


,



1
 −2 


ker(A) = span 
 1  .
0
9. (a) Consider an n × p matrix A and a p × m matrix B. Show that ker(B) ⊂ ker(AB), i.e., show that if
~x ∈ ker(B), then ~x ∈ ker(AB).
(b) Show that im(AB) ⊂ im(A), i.e., show that if ~y ∈ im(AB), then ~y ∈ im(A).
(a) If ~x ∈ ker(B), then B~x = ~0. Therefore, it must also be true that A(B~x) = AB~x = ~0, so ~x ∈ ker(AB).
(b) If ~y ∈ im(AB), then there exists ~x such that AB~x = ~y. But then A maps the vector B~x to ~y , so ~y ∈
im(A).
10. Are the vectors

 
 

1
1
1
 1 ,  2 ,  1 
1
3
6
linearly independent or linearly dependent? Justify your answer.
To see whether these vectors are linearly independent or dependent, we must check whether the linear relation





  
1
1
1
0
c1  1  + c2  2  + c3  3  =  0 
1
3
6
0
has only the solution c1 = c2 = c3 = 0, or whether we can satisfy the equation above
not equal to zero. The system above has the following coefficient matrix;

 
 
 
1 1 1
1 1 1
1 0 −1
1 0
A= 1 2 3 ∼ 0 1 2 ∼ 0 1 2 ∼ 0 1
1 3 6
0 2 5
0 0 1
0 0
with at least one of the ci

0
0 .
1
Since rref(A) = I, the only solution to the system A~c = ~0 is ~c = 0. This tells us that the given vectors are linearly
independent.
3