1. (5 points each) State whether the following are true or false. If false, give a
counterexample.
a. det(A + B) = det A + det B.
Solution:
1 0 0
False. Counterexample: A = 0 1 0 , B =


0 0 1
But det A = 1, det B = -4, and det (A + B) = 0.
 1 0 0
 0 2 0 . Then A + B =


 0 0 2
0 0 0 
0 3 0  .


0 0 3
b. An n×n matrix A has n pivot positions if dim( Nul A) = 0.
Solution:
True. Follows from the Invertible Matrix Theorem.
c. The set of all solutions of a system of m homogeneous equations in n unknowns is a
subspace of Rn.
Solution:
True. This is just a rephrasing of the definition of the null space of a matrix, which is a
subspace.
d. Suppose (B – C)D = 0 for suitable matrices B, C, and D, with D ≠ 0. Then B = C.
Solution:
1 2 
False. Counterexample: B = 
, C =
 2 4
0 0 
and (B – C)D = 
 ; but B ≠ C.
0 0 
1 2 
2 5 , D =


1 0
0 0 . Then B – C =


0 0 
0 1


1
2
2. Let A = 
3

5
2 3
0
 3 10  1
with A ~
 3 21  2

 7 27  3
1  2
0 1

0 0

0 0
3 0
4  1
.
0 1

0 0
a. (10 points) Find a basis for Col A. What is dim(Col A)?
Solution:
The basis for Col A is given by the pivot columns of A. Therefore:
1 
2
Basis (Col A) = {   ,
3
 
5 
 2
  3
 ,
  3
 
 7 
 0 
  1
  } and the dimension of Col A is 3.
 2
 
  3
b. (15 points) Find a basis for Nul A. What is dim(Nul A)?
Solution:
Put A into RREF to find a general solution to Ax = 0 and then write that solution in
parametric vector form to find the basis.
0 11 0
 11x3 
 11
  4x 
4
1 4 0
3 
.
. Thus a general solution is x = 
= x3 
 1 
 1 
0 0 1





0 0 0
 0 
 0 
 11
4
 }. It follows that dim(Nul A) = 1.
And a basis for Nul A is therefore { 
 1 


 0 
1
0
A~ 
0

0
c. (5 points) What does a solution to Ax = 0 look like?
Solution:
A general solution is just a linear combination of the vectors in the basis of Nul A. Thus,
 11a 
  4a 
 , for a  R.
a solution looks like x = 
 a 


 0 
d. (5 points) Is the linear transformation T: R4 → R4 defined by T(x) = Ax one-to-one?
Why or why not?
Solution:
A linear transformation is one-to-one if and only if the only solution to Ax = 0 is the
trivial solution. Since we have nontrivial solutions to Ax = 0 (given in part c.), this
transformation cannot be one-to-one.
a 
b 
3. (20 points) Suppose H = {   : c = a + 2b, d = a – 3b with a, b, c, d  R}. (i.e., H is
c 
 
d 
 a 
 b 
 ). Is H a subspace of R4?
the set of all vectors v in R4 such that v = 
a  2b 


 a  3b 
Solution:
Check the 3 properties for a subset to be a subspace:
(i) 0  H since we can let a = 0 and b = 0.
(ii) Given u, v  H, is u + v  H?
 a1 
 b 
1
,v=
Since u, v  H, u and v look like: u = 
a1  2b1 


 a1  3b1 
 a2 
 b

2

 (a1, a2, b1, b2  R).
a 2  2b2 


 a 2  3b2 
a1  a 2




b1  b2

  H, by definition of elements of H.
Then u + v =
(a1  a 2 )  2(b1  b2 )


 (a1  a 2 )  3(b1  b2 ) 
(iii) Given u  H, is cu  H, for c R?
 a 
 b 
 . Then cu =
Again, u  H looks like u = 
a  2b 


 a  3b 
which is an element of H by definition of H.
Therefore, yes, H is a subspace of R4.
ca
 ca  

 cb  

cb

 = 
,
c(a  2b) (ca )  2(cb)

 

 c(a  3b)  (c) a  3(cb) 
4. (15 points) Find a formula for det(rA) (in terms of det A) for an n×n matrix A and a
scalar r.
Solution:
Let A = [aij] and B = [bij]. Then rA = B, where bij = r aij.
Multiplying every entry of A by r is the same as multiplying every row of A by r.
Therefore, when we computer det(rA), we can factor out an r for every row of A. Since
A is an n×n matrix, there are n rows.
Thus, det(rA) = rndet A.
 cos( ) sin(  ) 
-1
5. (15 points) Show that A = 
 is invertible and find A .

sin(

)
cos(

)


Solution:
A is invertible if and only if det A ≠ 0.
So we compute det A = cos2() + sin2() = 1. Since this is nonzero, A is invertible.
Formula for A-1 =
Thus, A-1 =
1
1
1
det A
 d  b
 c a  .


cos( )  sin(  ) cos( )  sin(  )
 sin(  ) cos( )  =  sin(  ) cos( )  .

 

6. Given vectors a1 and a2 in R2:
DISCLAIMER: I did the best I could to graph these vectors… this is just roughly
where they should be, because LaTex is dumb and won’t let me do non-integer
slopes and stuff.
a. (5 points) Draw a1 + a2 on the figure above. Remember to label it.
 3 
b. (10 points) On the figure above, draw the image of x =   under the transformation
 2
T: R2 → R2 with standard matrix A = a1 a 2  . Remember to label it.
Solution:
x 
T is defined by T(x) = Ax = a1 a 2   1  = x1a1 + x2a2.
 x2 
 3 
Therefore, T(   ) = 3a1 – 2a2.
 2
 1
c. (10 points) If a1 =   and a2 =
2
determined by 0, a1, and a2?
5
1 , what is the area of the parallelogram
 
Solution:
The area of the parallelogram is the absolute value of the determinant of the matrix
whose columns are a1 and a2.
 1 5
Thus {Area of the parallelogram} = det 
 =  11 = 11.
 2 1
7. (20 points) Suppose S = { v1, v2, v3 } is a linearly independent set of vectors in R3. Is
T = { w1, w2, w3 }, where w1 = v1 + v2 ,w2 = v1 + v3 , w3 = v2 + v3, a linearly independent
set?
Solution:
A set of vectors { v1, v2, v3 } is linearly independent if and only if a1v1 + a2v2 + a3v3 = 0
implies a1 = 0, a2 = 0, and a3 = 0.
So we look at c1w1 + c2w2 + c3w3 = 0:
0 = c1w1 + c2w2 + c3w3 = c1(v1 + v2) + c2(v1 + v3) + c3(v2 + v3)
= (c1 +c2) v1 + (c1 +c3) v2 + (c2 +c3) v3
So we get the equation (c1 +c2) v1 + (c1 +c3) v2 + (c2 +c3) v3 = 0.
v1, v2, v3 linearly independent  (c1 +c2) = 0, (c1 +c3) = 0, and (c2 +c3) = 0.
We now have a system of linear equations – does it have a nontrivial solution?
1 1 0 0 1 1 0 0

 

To solve it, we look at the augmented matrix: A = 1 0 1 0 ~ 0  1 1 0
0 1 1 0 0 0 2 0
This matrix has a pivot in every column, which implies that the only solution to Ax = 0 is
the trivial solution. Therefore, c1 = 0, c2 = 0, and c3 = 0.
Thus { w1, w2, w3 }is a linearly independent set.
8. (Bonus problem, 5 points) Let A and B be n×n matrices. Show that if Ax = Bx for all
x  Rn, then A = B. (Hint: Since the statement is true for all x  Rn, what are some
“nice” vectors?)
Solution:
1 
0 
Since we have Ax = Bx for all x  Rn, let x = e1 =   .

 
0 
Then Ax = a1, where a1 is the first column of A. And Bx = b1, where b1 is the first
column of B.
0 

 
Similarly, we can let x = ei = 1  , where the 1 is in the ith row.
 

0 
Then we will have Ax = ai, where ai is the ith column of A. And Bx = bi, where bi is the
ith column of B.
So we have ai = bi for every i = 1, …, n. Since the columns of A are equal to the
corresponding columns of B, we have that the corresponding entries of each pair of
columns are also equal. Thus the entries of A are equal to the corresponding entries of B.
i.e., aij = bij for every i, j = 1, …, n. Therefore A = B.