SOLUTIONS TO PRACTICE EXAM 1
1. Multiple Choice
(1) Answer: (b)

   
 

1
2
0
1
(2) For which constants t do the vectors  1 ,  3 ,  −4 ,  −4  span
0
6
t
t
3
all of R ?. Solution: Consider the matrix


1 2 0 1
A =  1 3 1 −4  .
0 6 t t
The only way for the columns of A to span all of R3 is if the reduced echelon
form of A has pivots in every row. By means of row operations we obtain






1 2 0 1
1 2 0 1
1 2
0
1
2 −R1
3 −6R2
 1 3 1 −4  R2 →R
 0 1 1 −5  R3 →R
 0 1
1
−5  = A0 .
−→
−→
0 6 t t
0 6 t t
0 0 t − 6 t + 30
Even though the matrix A0 is not quite the reduced echelon form of the A,
it is clear from looking at A0 that the reduced echelon form of A has pivots
in rows 1 and 2. On the other hand, it is impossible for the third row of the
reduced echelon form of A to have all of its entries equal to 0 regardless of the
value of t, in fact, the only way for this to happen is if t − 6 and t + 30 vanish
simultaneously. It follows that the correct answer is (d), i.e., the vectors span
all of R3 for every t.


 
 
1
1
0
(3) Consider the vectors v1 =  −2 , v2 =  0 , v3 =  3 , v4 =
1
3
3


1
 −3 . The goal of the problem is to determine which subsets of {v1 , v2 , v3 , v4 }
0
are linearly dependent. Recall that a set of vectors is linearly dependent if
one of the vectors in the set is a linear combination of the rest of the elements
in the set.
A {v1 , v2 } are not linearly dependent (v1 and v2 are not multiples of each
other).
1
2
SOLUTIONS TO PRACTICE EXAM 1
B {v1 , v2 , v3 } are linearly dependent. Note that


0
2
v1 − v2 =  −2  = − v3 ,
3
−2
and hence v3 is a linear combination of v1 , v2 .
C {v1 , v2 , v4 } are linearly dependent. In fact, note that


−1
−3v1 + v2 = 2  3  = 2v4 .
0
D The set {v1 , v2 , v3 , v4 } is linearly dependent because it contains linearly
dependent sets. For example v4 = − 23 · v1 + 21 · v2 + 0 · v3 .
The answer is (c), i.e., B,C and D are the only linearly dependent sets.
(4) Which of the following matrices define a transformation which is one-to-one?


−1 −1 2
1 2 3
1 2
2 1 .
A=
B=
, C= 1
4 5 6
1 2
2
5 7
Solution: In order for a linear transformation to be one-to-one, it is necessary for its matrix to have linearly independent columns. Note that the set
of columns of A is a set of three vectors in R2 which can never be linearly
independent. It follows that A does not define a one-to-one linear transformation. The matrix B cannot define a one-to-one transformation either
because its columns are multiples of each other. Finally, C is row equivalent
to the matrix


1 0 0
 0 1 0 
0 0 1
which implies that its columns are linearly independent. C is then the only
matrix in the list that defines a one-to-one linear transformation. The correct
answer is (e).
(5) For which t is the matrix A below invertible


1 1 3
A =  2 2 2 ?
1 3 t
Solution: By using row operations


1 1 3
1
1 ,R3 →R3 −R1
 2 2 2  R2 →R2 −2R
 0
=⇒
1 3 t
0

we have



1 1
3
1
3
R2 ↔R3
 0 2 t−3 
0 −4  =⇒
2 t−3
0 0 −4
SOLUTIONS TO PRACTICE EXAM 1
and we can further reduce A to

1 1
 0 1
0 0
9
2
−
t−3
2
t
2
3

.
1
which gives us the same information as the reduced echelon form of A. It
follows that A is invertible because its reduced echelon form has pivots in every
column. Alternatively, one can solve this problem by using determinants.
Note that the determinant of A is
2 2
2 2
2 2
det(A) = det
− det
+ 3 det
3 t
1 t
1 3
= (2t − 6) − (2t − 2) + 3(6 − 2) = 8,
and therefore A is invertible for every
(e).

 t.Answer:
1
(6) Find the coordinates of the vector  2  relative to the basis B of R3 given
3
by
     
1
3 
 1





0 , 2 , 2  .
B=
 0
4
2 
 
1

2  with respect to the basis B are
Solution: The coordinates of v =
3
coefficients λ1 , λ2 , λ3 such that
   
 
 
3
1
1
1
 2  = λ1  0  + λ2  2  +  2  ,
2
3
0
4
in other words, the coordinates of v with respect to B are the components of
a column vector x in R3 such that
Ax = v,


−1
and solving this system we obtain x =  1/2 . And these are the coordi1/2
nates of v with respect to B. Answer: (a).
(7) Find the rank and dimension of the null space (nullity) of the matrix
1 3 5
A=
2 4 6
4
SOLUTIONS TO PRACTICE EXAM 1
Solution: Note that A is row equivalent to the matrix
1 0 −1
0
A =
0 1 2
which happens to be the reduced echelon form of A0 . It follows that A has
two pivot columns and therefore rank(A) = dim(Col(A)) = 2. On the
other hand we have the relation
rank(A) + null(A) = # of columns of A = 3,
and therefore null(A) = 1. Answer: (d).
(8) Find det(AB) where




1 5 6
1 0 1
A =  0 2 7 , B =  2 3 0 .
0 0 3
1 0 4
Observe that determinant is a multiplicative function for square matrices of
the same size and therefore det(AB) = det(A) det(B). On the other hand,
since A is upper triangular we have


1 5 6
det(A) = det  0 2 7  = 1 · 2 · 3 = 6,
0 0 3
and also
det (B) = det
3 0
0 4
+ det
2 3
1 0
= 12 − 3 = 9.
We conclude that det(AB) = 6 · 9 = 54. Answer: (c).
2. Partial Credit
(9) Find the solutions to the linear system below in parametric form
x1 + x3 + x4 = 1
x1 + 2x2 + x3 + 5x4 = 7
x1 + x2 + x3 + 4x4 = 3
(1)
Solution: Consider the augmented

1 0 1
A= 1 2 1
1 1 1
By means of row operations we

1 0 1
 0 1 0
0 0 0
matrix

1
7 
3
1
5
4
obtain
0
0
1

2
5 
−1
SOLUTIONS TO PRACTICE EXAM 1
5
Which means that system (1) is equivalent to
x1 + x3 = 2
x2 = 5
x4 = −1.
Observe that the general solution of (1) satisfies

 

 

x1
2 − x3
2
 x2  

  5 
5

 
 

x3 
 x3  =  x3  =  0  + |{z}

parameter
x4
−1
−1
|
| {z }
p
−1
0
1
0
{z
v


.

}
In the above description, p is a particular solution of the inhomogeneous system and x3 v is the general solution of the inhomogeneous system.
(10) Suppose that a linear transformation T : R2 → R3 satisfies


 
1
2
0
1



−2
1 
, T
=
T
=
1
0
1
0
(a) Find the standard matrix A of T .
Solution: We have


1 2
1
0
A= T
T
=  −2 1 
0
1
1 0


5
(b) Find a vector x so that T (x) =  5 .
−1
Solution: We have




1 2
5
1
1 0
Row operations
 −2 1
 0 1
5 
3 .
=⇒
1 0
−1
0 0
0
−1
It follows that x =
.
3
(c) Find a vector b in R3 which does not lie in the image of T (or explain
why there are none).
Solution: A sequence of row operations that take A to its reduced echelon form are R1 ←→ R3 , R2 → R2 + 2R1 , R3 → R3 − R1 , R3 → R3 − 2R2 .
6
SOLUTIONS TO PRACTICE EXAM 1


b1
It follows that for generic b =  b2  we have that
b3


b1
1 2
 −2 1
b2 
b3
1 0
is row equivalent to

1 0
 0 1
0 0

b3

b2 − 2b3
b1 − 2b2 + 3b3
and the only way b can be in the image of T is if
b1 − 2b2 + 3b3 = 0
(2)


1
Choose b =  0  which violates (2). It follows that b is not in the
0
image of T .
(11) Find the inverse of A, where


1 2 1
A =  1 3 1 .
1 4 0
Solution: Routine computation. The answer is


4 −4 1
0 .
A−1 =  −1 1
−1 2 −1
(12) Consider the matrix A given by
1 2 1 0
A=
3 7 4 −2
(a) Find a basis for the nullspace Nul(A). Solution: We need to find a basis
for the subspace of R4 consisting of all solutions of Ax = 0. The reduced
echelon form of A is
1 0 −1 4
0 1 1 −2
(3)
which implies that Ax = 0 is equivalent to
x1 − x3 + 4x4 = 0
x2 + x3 − 2x4 = 0.
SOLUTIONS TO PRACTICE EXAM 1
7
A basis of Nul(A) is obtained by setting x3 = 1, x4 = 0 and x3 = 0, x4 = 1
which leads to the vectors


 
−4
1









2
−1

,

 1   0 





1
0


14
 −8 

(13) The vector x = 
 2  is in Nul(A). Find the coordinates of x relative to
−3
the basis found in part (a).


x1
 x2 

Solution: From (3) we observe that any element x0 = 
 x3  satisfies
x4



 



−4
x3 − 4x4
1
x1




 x2   −x3 + 2x4 
=
 = x3  −1  + x4  2  .

 0 

 1 
 x3  
x3
1
0
x4
x4


14
 −8 

It follows that the coordinates of x = 
 2  with respect to the basis in
−3
2
.
part (a) are
−3