# Last name: name: 1 ```Last name:
name:
1
Mid term exam 1 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your
Question 1: Find a parametric representation of the line in R4 that passes through P =
(1, 2, 3, 4) and is orthogonal to the hyperplane x1 − 2x2 + 3x3 − 4x4 = 5.
A normal vector to the hyperplane x1 − 2x2 + 3x3 − 4x4 = 5 is u := (1, −2, 3, −4). The line in
question is the collection of points X = (x1 , x2 , x3 , x4 ) so that X − P is parallel to u. This means
that there is a real number t so that
X − P = tu = (t, −2t, 3t, −4t).
The parametric representation of the line is
x1 = 1 + t,
x2 = 2 − 2t,
x3 = 3 + 3t,
x4 = 4 − 4t.
Question 2: Find a unit vector orthogonal to u = (1, 2, 3) and v = (3, 2, 1).
It is a known property of the cross product that the following vector w := u &times; v is orthogonal to u
and v. By definition
2 2 3 1 1 3
)
(1)
,
,
w = (
3 1 1 3 2 2
= (−4, 8, −4).
By normalizing w we obtain the desired vector:
w
1
= √ (−4, 8, −4).
kwk
96
(2)
2
Mid term exam 1, September 28, 2010
Question 3: Find an equation of the hyperplane in R4 that passes through P = (0, 1, 2, 3) and
is normal to u = (−1, −1, 1, 1).
By definition H is the collection of points X = (x1 , x2 , x3 , x4 ) so that (X − P )&middot;u. This means
0 =(X − P )&middot;(−1, −1, 1, 1) = (x1 , x2 − 1, x3 − 2, x4 − 3)&middot;(−1, −1, 1, 1)
= −x1 − (x2 − 1) + (x3 − 2) + (x4 − 3) = −x1 − x2 + x3 + x4 − 4.
The equation of the hyperplane in question
0 = −x1 − x2 + x3 + x4 − 4.
Question 4: Prove the triangle inequality: ku + vk ≤ kuk + kvk for all u, v ∈ Rn .
By definition of the dot product, we have the following equality:
ku + vk2 := (u + v)2 = (u + v)&middot;(u + v) = u&middot;u + u&middot;v + v&middot;u + v&middot;u.
Using the Cauchy-Schwarz inequality, we obtain
ku + vk2 ≤ kuk2 + kukkvk + kvkkuk + kvk2 = (kuk + kvk)2 .
Taking the square root gives the triangle inequality:
ku + vk ≤ kuk + kvk.
Last name:
name:
Question 5: Find the inverse of A =
1
3
3
2
4
The determinant of A is |A| = 4 − 16 = −2 6= 0. The matrix is invertible. We use the formula
from class to compute the inverse:
1
4 −2
−1
A =
−2 −3 1
1 −1 1
1 2 3
B=
. Find (a) 3A − 4B (b) AT (c) AT B.
−1 1 −1
3 2 1
3 −3 3
−4 −8 −12
−1 −11 −9
3A − 4B =
+
=
−3 3 −3
−12 −8 −4
−15 −5 −7


1 −1
T

−1
1
A =
1 −1


−2 0 2
AT B =  2 0 −2
−2 0 2
Question 6: Let A =
4
Mid term exam 1, September 28, 2010
Question 7: Let A be a square matrix. Assume that A has the following property: X T AX ≥
2kXk2 for all column vector X. Prove that A is invertible.
Consider X so that AX = 0. Then 0 = X T AX ≥ 2kXk2 . This proves that kXk = 0, which in
turn proves that X = 0. As a result the solution set of AX = 0 is {0}. This proves that A is
invertible since A is square.
Question 8: Find 2 &times; 2 nonzero matrices A and B such that AB = 0.
Consider the following matrix:
A=
0
0
1
,
0
Then
AB = 0
B=A
Last name:
name:

5

−3
13 . Find the LU factorization of A.
−5
1
Question 9: Let A = −3
2
2
−4
1
We compute the echelon form

1
A −3
2
of A to get U and L
 
2 −3
1 2
−4 13  ∼ 0 2
1 −5
0 −3


1
l1 = −3 ,
2
 
−3
1
4  ∼ 0
1
0

0
l2 =  1  ,
− 23

2
2
0
 
0
l3 = 0
1
Finally we have

1
L = −3
2
0
1
− 32

0
0 ,
1

1
U = 0
0
2
2
0

−3
4
7

−3
4
7
6
Mid term exam 1, September 28, 2010
Question 10: Let u1 = (1, 2, 4), u2 = (2, −3, 1), u3 = (2, 1, −1) in R3 . (a) Show that u1 , u2 ,
u3 are orthogonal.
We compute ui &middot;uj for i 6= j, i, j ∈ {1, 2, 3}.
u1 &middot;u2 = 2 − 6 + 4 = 0,
u1 &middot;u3 = 2 + 2 − 4 = 0,
u2 &middot;u3 = 4 − 3 − 1 = 0.
This proves the statement.
(b) consider the 3 &times; 3 matrix U = [u1 u2 u3 ] where uj , j ∈ {1, 2, 3} are considered as column
vectors. Show that U is invertible.
 
x1
Consider X = x2  so that U X = 0. Then
x3
x1 u1 + x2 u2 + x3 u3 = 0.
Taking the dot product with u1 gives
0 = x1 u1 &middot;u1 + x2 u2 &middot;u1 + x3 u3 &middot;u1 = x1 ku1 k2 = 21x1 .
This proves that x1 = 0. By repeating this process with u2 and u3 we obtain that x2 = 0 and
x3 = 0. As a result the solution set of U X = 0 is {0}. This means that U is invertible.
(c) Let v = (3, 5, 2). Solve U X = v, where v is considered as column vector. (Do not compute
the reduced echelon form of the system.)
By definition
x1 u1 + x2 u2 + x3 u3 = v
Taking the dot product with u1 and using the orthogonality of u1 , u2 , u3 gives
v&middot;u1 = 21 = x1 u1 &middot;u1 + x2 u2 &middot;u1 + x3 u3 &middot;u1 = x1 ku1 k2 = 21x1 ,
i.e., x1 = 1. Similarly
v&middot;u2 = −7 = x1 u1 &middot;u2 + x2 u2 &middot;u2 + x3 u3 &middot;u2 = x2 ku2 k2 = 14x2 ,
i.e., x1 = − 21 .
v&middot;u3 = 9 = x1 u1 &middot;u3 + x2 u2 &middot;u3 + x3 u3 &middot;u3 = x3 ku3 k2 = 6x3 ,
i.e., x1 = 32 . Hence the solution is

1
X = − 12  .

3
2
```