Download MATH 51 MIDTERM 1 SOLUTIONS 1. Compute the following: (a). 1

MATH 51 MIDTERM 1 SOLUTIONS
1. Compute the following:


 
1 2 9
7



3 4
(a).
= 25 
1
−1 3
−4


7 0
15
−7
1 1 2 
4 1 =
(b).
4 −17
0 −1 4
2 −4
   
 
1
−1
1
(c). 32 + 2 1  = 8
5
−2
3
   
1
−1
(d). 2 ·  1  = −5
3
−2
2 2 6
1 3 =
(e).
7
7 21
1


1 0 3
2. Find the inverse of the matrix A = 1 1 5 .
3 2 14
Solution
We write the augmented matrix [A|I3 ]


1 0 3 1 0 0
1 1 5 0 1 0
3 2 14 0 0 1
and perform row operations until we get the identity on the left. In
the first step we do row2-row1 and row3-3row1 to get


1 0 3 1 0 0
0 1 2 −1 1 0
0 2 5 −3 0 1
then we do row3-2row2 and we get


1 0 3 1
0 0
0 1 2 −1 1 0
0 0 1 −1 −2 1
finally we do row1-3row and

1 0
0 1
0 0
row2-2row3 and we are left with

0 4
6 −3
0 1
5 −2
1 −1 −2 1
From here we deduce

4
6 −3
5 −2
= 1
−1 −2 1

A−1
.
2
3(a). Consider the following matrix A and its
form rref(A):



1
4 3 7 1 3
0
2 3 5 −1 2


A=
1 1 2 0 1 , rref(A) = 0
0
5 4 9 1 4
row reduced echelon
0
1
0
0

1 1 0
1 −1 0

0 0 1
0 0 0
(You do not need to check this.) Find a basis for the column space
C(B) of B.
Solution
If we take the columns of A correspondig to the position of the pivots
in rref(A) we get the basis
     
3
3 


 4

2 3 2
 ,  ,  
1 1 1



 5
4
4 
3
3(b). As in part (a),


4 3 7 1 3
2 3 5 −1 2

A=
1 1 2 0 1,
5 4 9 1 4

1
0
rref(A) = 
0
0
0
1
0
0

1 1 0
1 −1 0
.
0 0 1
0 0 0
Find some vectors that span the nullspace N (A) of A.
Solution
We want to find the x ∈ R5 that satisfy Ax = 0. For this we use
rref(A) and get the equations
x1 = −3x3 − x4
x2 = −x3 + x4
x5 = 0
This implies that we can write x as
 
 
 
−1
−1
x1
1
−1
 x2 
 
 
 
 x3  = x 3  1  + x4  0 
 
 
 
1
0
 x4 
0
0
x5
and therefore a basis for N (A) is the set
   

−1
−1 









−1
   1 

 1 ,  0 
   


   1 


 0


0
0 
4


1 3 0
4(a). Consider the equation Ax = b where A = 2 7 0. Find the
1 2 0
condition(s) on the vector b for this equation to have a solution. (Your
answer should
 be one or more equations involving the components bi
b1
of b = b2 .)
b3
Solution
We write the augmented matrix


1 3 0 b1
2 7 0 b2 
1 2 0 b3
and proceed to row reduce it. First we do row2-2row1 and row3-row1
and get


1 3 0
b1
0 1 0 b2 − 2b1 
0 −1 0 b3 − b1
bext we do row3+row2 to get


1 0 0 b1 − 3(b2 − 2b1 )

0 1 0
b2 − 2b1
0 0 0 b3 − b1 + b2 − 2b1
and for the system to have a solution we must have
b3 − b1 + b2 − 2b1 = b2 + b3 − 3b1 = 0
5
4(b). Find a matrix B such that N (B) = C(A), where


1 3 0
A = 2 7 0.
1 2 0
(This is the matrix from part (a).)
Solution
First we note that by part a) we have

 

 b1
C(A) = b2  ∈ R3 / − 3b1 + b2 + b3 = 0


b3
We might rewrite the equation −3b1 + b2 + b3 as
 
b1
−3 1 1 b2  = 0
b3
and from this we can tell that the matrix B = −3 1 1 will satisfy
N (B) = C(A).
6
3
5(a). Find all solutions of the equation Ax =
, where
16
1 −2 2 1
A=
.
5 −10 11 2
Solution
We write the augmented matrix
1 −2 2 1 3
5 −10 11 2 16
and proceed to row reduce it. First we do row2-5row1 to get
1 −2 2 1 3
0 0 1 −3 1
1 −2 0 7 1
0 0 1 −3 1
and then row1-2row2
The equivalent system of equations is now
x1 = 1 + 2x2 − 7x4
x3 = 1 + 3x4
From here we deduce that the set of solutions is
 
 
 
1
2
−7 



 
1
 0 
0
  + x2   + x4  
1
0
 3 



 0
0
1 
.
7
5(b). Find a basis for the nullspace of A, where A is the matrix in
part (a).
Solution
From part a) we get the basis
   
2
−7 



   
1
 ,  0 
0  3 



 0
1 
8
6(a). Suppose S is the sphere in R3 of radius 1 centered at the origin.
Suppose that A, B, and C are three points on S and that side AC of
the triangle ABC is a diameter of the sphere. Prove using vectors that
the triangle has a right angle at B.
Solution
If we let A = (a1 , a2 , a3 ), then since AC is a diameter we must have
C = −A = (−a1 , −a2 , −a3 ). Let B = (b1 , b2 , b3 ). To prove that the
triangle ABC has a right angle at B is the same as proving that the
−→
−−→
vectors AB = B − A and CB = B − C are perpendicular. Now we
compute
−→ −−→
AB · CB = (B − A) · (B − C)

 

b 1 − a1
b 1 + a1
= b2 − a2  · b2 + a2 
b 2 − a3
b 3 + a3
= (b1 − a1 )(b1 + a1 ) + (b2 − a2 )(b2 + a2 ) + (b3 − a3 )(b3 + a3 )
= b21 − a21 + b22 − a22 + b23 − a23
= (b21 + b22 + b23 ) − (a21 + a22 + a33 )
And since A and B are points in the sphere we have b21 + b22 + b23 =
−→ −−→
a21 + a22 + a23 = 1, and that implies AB · CB = 0.
9
6(b). Suppose A is a matrix with 4 columns. Suppose u and v are
linearly independent vectors in R4 such that Au = Av = 0. Prove that
if x, y, and z are any vectors in R4 , then the vectors Ax, Ay, and Az
must be linearly dependent.
Solution
We can rephrase teh fact that Au = Av = 0 by saying that u, v ∈
N (A). By condition of the problem u and v are linearly independent
and that implies dimN (A) ≥ 2. A has 4 columns, therefore by proposition 12.4 (page 77 in the book) we must have rank(A) + nullity(A) = 4
and this implies nullity(A) ≤ 2. Since the dimension of the column
space is at most 2 any three vectors must be linearly dependent.
10
7(a). Find a parametric equation for the line L through the points
A = (−1, 0, 3) and B = (2, 1, 5).
Solution
First we find the direction vector v = B − A = (3, 1, 2). Now the
parametric equation of the line is
   
 
x1
−1
3
x2  =  0  + t1
x3
3
2
11
7(b). Find a point C on L such the line OC and the line L are
perpendicular to each other. (Here L is the line from part (a), and the
line OC is the line through the origin O = (0, 0, 0) and C.)
Solution
If we look at the equation we found in part a), any point of the line is
given by a value of t, let us assume that the point C is given by


−1 + 3t
C= t 
3 + 2t
The line OC has direction C and the line L has direction (3, 1, 2).
In order for two lines to be perpendicular this two vectors must be
perpendicular, ie. we must have
  

3
−1 + 3t
 t  · 1 = 0
2
3 + 2t
−3 + 9t + t + 6 + 4t = 0
14t = −3
3
t = −
14
3 36
23
And from here we find C = (− 14 , − 14 , 14 ).
12
8. Find a matrix A such that:
 
 
1
3
3
7
A1 =
, A1 =
,
1
13
1
1
 
3
7
A2 =
.
18
1
Solution
 
 
 
1
3
3
Let us name the vectors v1 = 1, v2 = 1 and v3 = 2. We can
1
1
1
find e1 = 21 (v2 − v1 ) and e2 = v3 − v2 . This implies
1
1 4
2
(Av1 − Av2 ) =
Ae1 =
=
6
2
2 12
0
Ae2 = Av3 − Av2 =
5
We notice that e3 = v1 − e1 − e2 and using the above results we get
1
Ae3 = Av1 − Ae1 − Ae2 =
−10
Finally, Ae1 , Ae2 and Ae3 are the columns of A, therefore we obtain
2 0 1
A=
6 5 −10
13
9(a). Suppose ∆ is an equilateral triangle in R3 each of whose sides
has length 7. Let A, B, and C be the corners of ∆. Find
−→ −→
2 AB · 3 AC .
−→
[Notation: AB is the vector beginning at A and ending at B.]
Solution
We do the computation
−→ −→
−→ −→
2 AB · 3 AC
= 6 AB · AC
−→ −→
= 6||AB||||AC|| cos θ
−→
−→
−→
−→
Where ||AB|| = ||AC|| = 7, and θ = π/3 is the angle that AB and AC
make (visually, to find the angle between the vectors we need to draw
them starting from the same point). Thus, we find that the answer is
(6)(7)(7)( 21 ) = 147 .
14
9(b). Suppose u, v, and w are unit vectors in R4 such that each one is
orthogonal to the other two. Find a number c so that u + 2v + 3w and
5u + v + cw are orthogonal to each other. (Recall that a unit vector
is a vector whose norm is 1.)
Solution
The problem is equivalent to find c such that (u + 2v + 3w) · (5u + v +
cw) = 0. Now we have
(u + 2v + 3w) · (5u + v + cw) = 5u · u + u · v + cu · w
+10v · u + 2v · v + 2cv · w
+15w · u + 3w · v + 3cw · w
= 5||u||2 + 2||v||2 + 3c||w||2
Here we used the fact that the vectors u, v and w are orthogonal, ie.
u · v = u · w = v · w = 0. Now we use the fact that they have norm 1
and the equation reduces to 5 + 2 + 3c = 0, therefore c = − 73 .
15
10. Short answer questions. (No explanations required.)
(a). Suppose that a linear subspace V is spanned by vectors v1 , v2 , . . . , vk .
What, if anything, can you conclude about the dimension of V ?
Answer: dimV ≤ k. Since we have k vectors spanning V a set with
more than k vectors must be linearly dependent (proposition 12.1 in
the book). Since a basis consists of linearly independent vectors we
must have dimV ≤ k
(b). Suppose that a linear subspace W contains a set {w1 , w2 , . . . , wk }
of k linearly independent vectors. What, if anything, can you conclude
about the dimension of W ?
Answer: dimW ≥ k. If we had a basis with less than k elements, again
by proposition 12.1 a set of k vectors could not be linearly independent.
(c). Suppose u · v > 0. What, if anything, can you conclude about the
angle θ between u and v? [Note: by definition, the angle θ between
two nonzero vectors is in the interval 0 ≤ θ ≤ π.]
Answer: 0 ≤ θ < π2 . Since u · v = ||u||||v||cosθ and u · v > 0 we must
have cosθ > 0.
(d). Suppose that A is a k × m matrix (i.e., a matrix with k rows and
m columns.) If k < m, what, if anything, can you conclude about the
number of solutions of Ax = b?
Answer: Either there is no solution or if a solution exist there are
infinitely many. This is because according to proposition 12.4 we have
rank(A)+nullity(A) = m, and since the columns of A are vectors in Rk
we have rank(A) ≤ k, therefore nullity(A) > 0. If b is in the column
space of A we will have infinitely many solutions, otherwise we will
have no solution.
(e). Suppose A is a matrix with 5 rows and 4 columns. Suppose that
the equation Ax = 0 has only one solution. What, if anything, can you
conclude about the dimension of the column space of A?
Answer: dimC(A) = 4. The only solution for Ax = 0 must be x = 0
and therefore N (A) = 0. Again we obtain the answer using proposition
12.4.
16