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Prof. Gabel
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Math 203 Matrix Algebra -- Exam I: Professor Gabel (March 4, 2004)
Directions: Do ALL of your work on these sheets of paper. Show all your steps. No calculators are allowed.
Remember, the honor code is to be observed on this exam. You are allowed one study sheet on this exam.
Suggestion: some of the problems say: show something. Be guided in your answer by how much space I have
left for your work. In particular, nearly all of these problems require little space to solve. In order to maximize
the credit you receive, show all of your steps, write neatly and give some reasons. Think a little, think a bit
more, and then start writing. Problems marked NPC are No Partial Credit.
There are 150 points on this exam Make sure you have 4 pages and put your name on each sheet.
(1) NPC Each of the questions below is worth 3 points (making 60 points total). Circle T for True, F for False
or circle blank if you wish to leave the problem blank.
SCORING: correct = +3, blank = +1 wrong = +0
A reminder on the use of some words: Consider the statement.
Let x and y be any whole numbers, each >1.
Then
"x+y must be > 1"
"x+y might be >1"
is TRUE, because as long as both x and y are > 1, then, of course, x+y >1
is TRUE, because if something MUST be true, then it MIGHT be true.
"x+y must be even"
"x+y might be even"
is FALSE, because it's not ALWAYS true. For example, if x=2 and y=3, then x+y is not even (it's odd).
is TRUE, because x could be 3 and y could be 7 and in this case, x+y=10, which is even.
"x+y must be < 0"
"x+y might be < 0"
is FALSE, because its not ALWAYS true For example, if x=2 and y=6 then x+y=8 which is not < 0.
is FALSE because there are no x and y both > 1 whose sum is negative.
Convention: if something must be true then it might be true.
T F blank Assume A is row equivalent to B. If Av=0 then it MUST be that Bv=0.
T F blank Assume A is row equivalent to B. If Av=0 then it MIGHT be that Bv=0.
T F blank
T F blank
Let A be a 3x2 matrix. Then A MUST have a right inverse.
Let A be a 3x2 matrix. Then A MIGHT have a right inverse.
T F blank
T F blank
If the columns of an mxn matrix span Rm then the rows MUST be independent.
If the columns of an mxn matrix span Rm then the rows MIGHT be independent.
T F blank
T F blank
If vectors v1, … vp span Rm, then p MUST be > m.
If vectors v1, … vp span Rm, then p MIGHT be > m.
T F blank
T F blank
If vectors v1, … vp in Rm are independent, then p MUST be < m.
If vectors v1, … vp in Rm are independent, then p MIGHT be < m.
T F blank
T F blank
If the columns of an nxn (square) matrix are independent, then they MUST span Rn.
If the columns of an nxn (square) matrix are independent, then they MIGHT span Rn.
Page 1 of 4
Prof. Gabel
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T F blank If the {v1, v2, v3} are independent and w is a 4th vector, then {v1, v2, v3, w} MUST be independent.
T F blank If the {v1, v2, v3} are independent and w is a 4th vector, then {v1, v2, v3, w} MIGHT be independent.
T F blank
T F blank
If a matrix A has an inverse, then the inverse of A MUST have an inverse.
If a matrix A has an inverse, then the inverse of A MIGHT have an inverse.
T F blank
T F blank
If v1, v2, w are vectors in R3 with Span{v1, v2} a plane, then Span{v1, v2, w} MUST be a plane.
If v1, v2, w are vectors in R3 with Span{v1, v2} a plane, then Span{v1, v2, w} MIGHT be a plane.
T F blank
T F blank
If A is a square matrix with a pivot in each row, then A MUST have a pivot in each column.
If A is a square matrix with a pivot in each row, then A MIGHT have a pivot in each column.
(2) NPC [10 pts] Consider the (true) statement concerning various products: (AB)v = A(Bv), where A and
B are matrices and v is a vector. Notice that there are four 's. Circle the correct responses:
Moving from left to right:
the first  is:
[(matrix)(matrix) multiplication] OR [(matrix)(vector) multiplication]
the second  is
[(matrix)(matrix) multiplication] OR [(matrix)(vector) multiplication]
the third  is
[(matrix)(matrix) multiplication] OR [(matrix)(vector) multiplication]
the fourth  is
[(matrix)(matrix) multiplication] OR [(matrix)(vector) multiplication]
the "=" sign is
[equality of vectors]
 1 4
 and let B =
(3) NPC [6 pts] Let A = 
 2 5
OR [equality of matrices]
 1  1

 . Then:
0 2 
 1 4   1  1  1 7 
 
 = 

(a) AB = 
 2 5 0 2   2 8
 1 4   3 12 
 = 

(b) 3A = 3 
 2 5   6 15 
1 2 0


(4) NPC [10 pts] Let A =  1 3 0  . Find its inverse. [I suggest you check your answer.]
0 0 1


 1 0 0 3  2 0


Row reduce (A | I3 )and if you do you get (I|A ) =  0 1 0  1 1 0 
0 0 1 0
0 1 

-1
Page 2 of 4
Prof. Gabel
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(5) NPC [12 pts] Consider the system:
x1 + 2x2 - 4x3
-3x6 = 5
-2x6 = 4
x5 + x6 = 8
Next to the free variables, put the word "free" and next to the basic (non-free) variables, put the actual formula
for them in terms of the free variables and constants. [Be careful about MINUS signs!]
x4
Variable
x1 =
x2 =
x3 =
x4 =
x5 =
x6 =
"free" or a formula
-2x2 + 4x3 +3x6 +5
free
free
2x6 +4
-x6 +8
free
(6) NPC Assume there is this linear system with unknowns x1, x2, x3, x4, x5 and assume further
Variable
"free" or formula
x1 =
-3x2 + x3 -4x5 +2
x2 =
free
x3 =
free
x4 =
-2x5 -7
x5 =
free
(a) [12 pts] Write the solution to this system in the format below: [Be careful about MINUS signs!]
  3
 1
  4  2 
 
 
   
 1 
 0
 0   0 
x 2  0   x3  1   x5  0    0 
 
 
   
 0
 0
  2   7 
 0
 0
 1   0 
 
 
   
(b) [8 pts] If this system were written in the format Ax=b, then what are A and b? [Be careful about MINUS signs!]
 1 3 1 0 4
 b =
A = 
 0 0 0 1 2
 2 
 
  7
1 3 5 
 .
(7) NPC [5 pts] Find the reduced echelon form of the matrix: 
 2 8 14 
 1 3 5   1 0  1

  

 2 8 14   0 1 2 
Page 3 of 4
Prof. Gabel
Sign Name: _________________________Print Name_________KEY_____________________
(8) [5 pts] Given v1, … vp, recall that Span {v1, … vp} = {a1v1+…+apvp | each ai is a real number}. Use this definition to show
 1   0 
 2
  is an element of Span  ,  .
6
 0   3 
 2
 1
 0
  (2)   (2)  , by defintion of Span, we are done.
 6
 0
 3
Since 
(9) NPC [7 pts] Let T: R3  R2 be the transformation given by T(x, y, z) = (x+2y, x+3z). Find the standard matrix for T.
Your Answer =
1 2 0 

 [Recall, you just compute T(e1), T(e2) and T(e3) and put those columns into a matrix.
1 0 3 
(10) [5 pts] Recall: A collection of vectors, v1, v2, ..., vn, is said to be linearly independent if the only linear combination of them that
is zero is the trivial linear combination. That is, if x1v1 + x2v2 +...+ xnvn = 0, then it must be that x1=x2=.... =xn=0. Using this
 1  4
 7
   
 
definition, show that  2  ,  0 , and  6  are NOT linearly independent:.
 2  0
 6
   
 
1
 
Since (3)  2  + (1)
 2
 
 4
 
 0  +(-1)
 0
 
7
 
 6 =
 6
 
2
Explain.
 0
 
 0  , by defintion, we are done.
 0
 
 x
 
(11) [10 pts] Consider T: R  R given by T  y  =
z
 
3
[Hint: You can do the relevant computation in your head.]
 x
 
 x
(a) Show that T is not one-to-one by finding vectors v and w in R3 with vw and yet T(v)=T(w). Show your answer is correct.
[That is, just don't simply say what v and w are, but say "vw" and compute T(v) and T(w) and show they are different.]
NOTE: I should have said THE SAME instead of ARE DIFFERENT, so I gave everyone full credit for this problem.
 0
 0
 
 
However, here is the solution anyway: Let v=  0  and w=  0  . Then clearly v  w and yet T(v)=T(w)=
1
 0
 
 
 0
 
 0 .
 0
 
[Note, lots of v and w work.]
(b) Show that T is not onto by finding a vector in R2 that is not the image under T of any vector in R3. Show your answer is correct.
[That is, explain why the vector in R2 that you found cannot be the image under T of any vector in R 3. Use some words.]
Clearly
1
  is not in the image of T as all vectors in the image of T have the same entry in each position.
 0
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