Math 317
Exam 1
Name:
Directions: Please answer all questions. Pen or pencil only. To receive full credit all work or necessary
justifications must be given.
1. Find the reduced row echelon form and the rank

1 2 3
 2 4 5
2 4 6
of the matrix

1
1 .
2
Solution:

1
 2
2
2
4
4
3
5
6

1
1 
2

1 2
 0 0
0 0

3
1
−1 −1 
0
0
The rank is 2.
1

1
 0
0
2
0
0
3
1
0

1
1 
0

1
 0
0
2
0
0
0
1
0

−2
1 .
0
2. Consider the system
x1 + 2x2 + 3x3 = 1
2x1 + 4x2 + 5x3 = 1
2x1 + 4x2 + 6x3 = 2.
a. Write a vector equation that is equivalent to the given system of equations:





 

1
2
3
1
x1  2  + x2  4  + x3  5  =  1 
2
4
6
2
b. Write a matrix equation that is equivalent

1 2
 2 4
2 4
to the given system of equations.

 

3
x1
1
5   x2  =  1  .
6
x3
2
c. Find the solution set of the given system of equations.
Solution: Consulting the reduced row echelon form in problem 1, we see that the given system is
equivalent to the system
x1 + 2x2 = −2
x3 = 1.
So every point (x1 , x2 , x3 ) in the solution set must satisfy the constraints x1 = −2x2 −2, x3 = 1, x2 ∈ R.
Thus the solution set is
{(−2x2 − 2, x2 , 1) : x2 ∈ R} = {(−2, 0, 1) + x2 (−2, 1, 0) : x2 ∈ R}.




1
1
d. If possible, express the vector b =  1  as a linear combination of the vectors v1 =  2  , v2 =
2
2
 


2
3
 4  , v3 =  5 .
4
6
Solution: Yes, it’s possible to express b as a linear combination of v1 , v2 , v3 because the system is
consistent. Choosing any (x1 , x2 , x3 ) in the solution set found in part c. and plugging these values into
the equation in part a. gives a sufficient answer.
For example, (−2, 0, 1) is in the solution set. So
b = −2v1 + 0v2 + 1v3 = −2v1 + v3
is a valid linear combination.
2
3. Let T : R2 → R2 be the linear transformation that rotates vectors counter-clockwise by π/3 radians
and let S : R2 → R2 be the linear transformation that reflects vectors across the line x2 = x1 .
a. Write the standard matrix of T .
A=
cos π/3
sin π/3
"
− sin π/3
cos π/3
0
1
1
0
0
1
1
0
3
2
1
2
#
=
√
1
√2
3
2
−
#
3
2
1
2
.
b. Write the standard matrix of S.
B=
.
c. Write the standard matrix of T ◦ S.
"
AB =
√
1
√2
3
2
−
3
2
1
2
#
"
=
√
−
3
2
1
2
1
√2
3
2
#
1
2√
#
d. Write the standard matrix of S ◦ T .
BA =
0
1
1
0
"
1
√2
3
2
3
√
−
"
=
√
3
2
1
2
−
3
2
4. Let a = (1, 2, 3) and consider the plane in R3 given by the equation a · x = −4. Find the distance from
the origin to the closest point on this plane.
Solution: The distance we’re interested in is precisely kproja xk for any x that satisfies the equation
a · x = −4. Observe that
|a · x|
a·a
4
√ .
kproja xk =
2 a = kak =
14
kak
4
5. Complete the definition:
a. A linear combination of the vectors v1 , . . . , vk is...
a vector of the form c1 v1 + · · · ck vk where c1 , . . . , ck ∈ R.
b. The span of the vectors v1 , . . . , vk is...
the set of all linear combinations of v1 , . . . , vk .
c. A linear transformation T : Rn → Rm satisfies the following two properties...
1. T (u + v) = T (u) + T (v) for all u, v ∈ Rn .
2. T (cv) = cT (v) for all v ∈ Rn and all c ∈ R.
5
6. Answer the following true/false questions and provide a brief justification or counterexample.
a. If A, B, C are 2 × 2 matrices with AB = CB, then A = C.
False: Here’s a counterexample:
A=
1
1
1
1
,B =
0
0
0
0
,C =
2
2
2
2
.
b. If the system Ax = b has two different solutions then it has infinitely many.
True: Let u, v be solutions of Ax = b. Let p ∈ R. Then observe that pu + (1 − p)v is a solution as
well since
A(pu + (1 − p)v) = pAu + (1 − p)Av = pb + (1 − p)b = b.
So we’ve shown that there are (at least) as many solutions as there are real numbers.
c. The standard matrix of a linear transformation T : R2 → R3 is a 3 × 2 matrix.
True: Since the standard matrix A of T must satisfy Ax = T (x) for all x ∈ R2 , A must have two
columns so that the product Ax is defined. Moreover, since T (x) ∈ R3 , A must have three rows so
that Ax ∈ R3 .
d. Let A be an m × n matrix. If rank A = m and Ax = 0 has a unique solution then Ax = b has a unique
solution for every b ∈ Rn .
True: Since rank A = m, the system Ax = b has a solution for every b. Fix such a b. Suppose
Ax = b has two solutions u1 , u2 . Then
A(u1 − u2 ) = Au1 − Au2 = b − b = 0,
which implies u1 − u2 = 0 since Ax = 0 is assumed to have only the trivial solution. Thus u1 = u2 .
e. If T is a linear transformation then T (0) = 0.
True: Let x be in the domain of T . Then
T (0) = T (x − x) = T (x) − T (x) = 0.
6