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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
answer. Answers with no justification will not be graded.
Question 1: Find a parametric representation of the line, L, in R4 that passes through P =
(1, 2, 0, 4) in the direction of u = (4, 0, 2, 1).
The line in question consists of the points X = (x1 , x2 , x3 , x4 ) in R4 that are such that there is
t ∈ R so that (X − P ) = tu. This means that
X − P = tu = (4t, 0t, 2t, t).
The parametric representation of the line is
x1 = 1 + 4t,
x2 = 2 + 0t,
x3 = 3 + 2t,
x4 = 4 + t.
or
L = {(1 + 4t, 2, 2t, 4 + t) ∈ R4 | t ∈ R}.
Question 2: Find an equation of the hyperplane, H, in R3 that passes through P = (1, 2, −3)
and is orthogonal to the vector n = (3, 2, −1).
By definition H is the collection of points X = (x1 , x2 , x3 ) so that (X − P )·n = 0. This means
0 =(X − P )·(3, 2, −1) = (x1 − 1, x2 − 2, x3 + 3)·(3, 2, −1) =
= 3x1 − 3 + 2x2 − 4 − x3 − 3 = 3x1 + 2x2 − x3 − 10.
The equation of the hyperplane in question is
3x1 + 2x2 − x3 = 10,
or
H = {(x1 , x2 , x3 ) ∈ R3 | 3x1 + 2x2 − x3 = 4}.
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Mid term exam 1, October 9, 2012
T
Question 3: A complex square matrix M is said to be
if M
which
√ unitary
M = I. Determine
−i
−
3
1
−
i
1
+
i
of the following matrices are unitary: A = 12 √
, B = 12
.
1+i 1−i
3
i
T
We compute AA ,
T
AA =
1 −i
√
3
4
√ − 3
i
√
i
− 3
√ 1
3
=
0
−i
0
1
T
This proves that A is unitary. We compute BB , (using that (1+i)(1−i) = 2 and (1+i)2 +(1−i)2 =
0)
1 1−i 1+i 1+i 1−i
T
1 0
=
BB =
0 1
4 1+i 1−i 1−i 1+i
This proves that B is unitary.

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Question 4: (a) Compute the echelon form of the augmented matrix M = 2
6

2
M = 2
6
2
2
6
1
3
4
1
5
6
 
2
3
5  ∼ 0
0
10
2
0
0
1
2
1
 
3
2
2 ∼ 0
4
0
1
4
3
2
0
0
1
2
0
1
4
1
2
2
6
1
3
4
1
5
6

3
5
10

3
2
0
(b) Compute the reduced echelon form of M .

2
M ∼ 0
0
2
0
0
1
2
0
1
4
1
 
3
2
2 ∼ 0
0
0
2
0
0
1
1
0
1
2
1
 
3
2
1 ∼ 0
0
0
2
0
0
The reduced echelon form of M is

1
M ∼ 0
0
1
0
0
0
1
0
0
0
1

1
1
0
0
1
0
−1
2
1
 
2
2
1 ∼ 0
0
0
2
0
0
0
1
0
0
0
1

2
1
0
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Question 5: Find a 2×2 real-valued orthogonal matrix P whose first row is a multiple of
(2, −2).
Recall that a real-valued matrix is orthogonal iff its rows (or columns) form an orthonormal set. The
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first row of P is λ(2, −2)T . Since this row is orthonormal, λ is such that |λ|(4 + 4) 2 = 1. This
1
1
means λ = ±8− 2 , and the first row of P is ±2− 2 (1, −1). The second row of P must be orthogonal
to the first one; this implies that it is proportional to (2, 2). This row must also be a unit vector;
1
this means that it is equal to ±2− 2 (1, 1). In conclusion there are four possibilities:
1 −1
− 12 1 −1
− 21
P = ±2
or P = ±2
.
1 1
−1 −1
−6x3 + 2x1 + 2x2 = 1
3x2 + 4x1 = 8
Question 6: Consider the following linear system
2x2 = 4
(a) Write the augmented matrix.
The system can be rewritten into the following form:
2x1 + 2x2 − 6x3 = 1
4x1 + 3x2 + 0x3 = 8
0x1 + 2x2 + 0x3 = 4
and the augmented matrix is

2
4
0
2
3
2
−6
0
0

1
8
4
(b) Determine the pivots and free variables.
We reduce the augmented

2 2
4 3
0 2
matrix in echelon form.
 
−6 1
2 2 −6
0 8 ∼ 0 −1 12
0 4
0 2
0
 
1
2
6 ∼ 0
4
0
2
−1
0
−6
12
24

1
6
16
There is no free variable. There are three pivots, which are 2, −1 and 24 in the first, second and
third column, respectively.
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Mid term exam 1, October 9, 2012


−3
13 . Find the LU factorization of A.
−5
1
Question 7: 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

−3
4
7
 
0
l3 = 0
1
Finally we have

1
L = −3
2
0
1
− 32

0
0 ,
1

1 2
U = 0 2
0 0

−3
4
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Question 8: Show that S = {1, 1−X, (1+X)2 } is a basis of the vector space P2 (X) (polynomials
of degree at most 2 over the field R).
Let us show that {1, 1 − X, (1 − X)2 } is linearly independent. Assume that there are three real
numbers x1 , x2 , x3 so that
x1 + x2 (1 − X) + x3 (1 + X)2 = 0,
then
x1 + x2 + x3 + (−x2 + 2x3 )X + x3 X 2 = 0.
The above polynomial is zero iff its coefficients are zero,
x1 + x2 + x3 = 0,
−x2 + 2x3 = 0 x3 = 0.
This implies immediately that x1 = x2 = x3 = 0, thereby proving that {1, 1 − X, (1 + X)2 } is
linearly independent. We know that the dimension of P2 (X) is 3 (since {1, X, X 2 } is a basis of
P2 (X)). As a result {1, 1 − X, (1 + X)2 } is also a basis of P2 (X) since this linearly independent
set contains 3 elements.
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1
1
1
,
1
1
5
−1
1
,
0
0
−1
1
,
0
0
0
is a basis of the
0
4 −1
vector space of 2×2 matrices with real coefficients. Find the coordinate vector of A =
2 1
relative to S.
We write A as a linear combination of the basis vectors in S. There are four real numbers x1 , x2 ,
x3 and x4 so that
1 1
1 −1
1 −1
1 0
A = x1
+ x2
+ x3
+ x4
.
1 1
1 0
0 0
0 0
Question 9: Accept as a fact that S =
This implies
4
2
−1
x + x2 + x3 + x4
= 1
1
x1 + x2
x1 − x2 − x3
x1
which in turn implies that
x1 = 1,
x1 + x2 = 2,
x1 − x2 − x3 = −1,
x1 + x2 + x3 + x4 = 4.
We then have x1 = 1, x2 = 1, x3 = 1, x4 = 1; as result, the coordinate vector of A relative to S is
[A]S = (1, 1, 1, 1)T .
Question 10: Let U be a n×n matrix with coefficients in R and assume that the columns of
U are nonzero and orthogonal in Rn . Prove that U is invertible.
It is sufficient to prove that the null space of U is zero, since U is a square matrix. Let C1 , C2 , . . . ,
Cn be the columns of U . Let X = (x1 , . . . , xn ) be a member of N (U ), i.e., N X = 0, then
U X = [C1 C2 . . . Cn ]X = x1 C1 + x2 C2 + . . . + xn Cn = 0.
Taking the dot product of this equation with C1 we obtain
x1 C1 · C1 + x2 C2 · C1 + . . . + xn Cn · C1 = 0.
Since the columns of U are orthogonal C2 · C1 = 0, . . . Cn · C1 = 0, we conclude that
x1 kC1 k2 = 0.
This means x1 = 0 since C1 is nonzero. Proceed similarly for x2 , . . . , xn . This shows that X = 0,
i.e., N (U ) = {0}, thereby proving that U is invertible.
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Mid term exam 1, October 9, 2012
Question 11: Let C1 = (2, 1, 4)T , C2 = (−3, 2, 1)T , C3 = (1, 2, −1)T in R3 . (a) Show that C1 ,
C2 , C3 are orthogonal.
We compute Ci ·Cj for i 6= j, i, j ∈ {1, 2, 3}.
C1 ·C2 = −6 + 2 + 4 = 0,
C1 ·C3 = 2 + 2 − 4 = 0,
C2 ·C3 = −3 + 4 − 1 = 0.
This proves the statement.
(b) Consider the 3×3 matrix with columns C1 , C2 , C3 , i.e., U = [C1 C2 C3 ]. Solve U X = V ,
where V = (10, 6, 4)T . (Do not compute the echelon form or the reduced echelon form of the
system. Use only the fact that the columns of U are orthogonal.)
Let X = (x1 , x2 , x3 )T ∈ R3 be a the solution of AX = V . By definition
V = x1 C1 + x2 C2 + x3 C3
Taking the dot product with C1 and using the orthogonality of C1 , C2 , C3 gives
V ·C1 = 42 = x1 C1 ·C1 + x2 C2 ·C1 + x3 C3 ·C1 = x1 kC1 k2 = 21x1 ,
i.e., x1 = 2. Similarly
V ·C2 = −14 = x1 C1 ·C2 + x2 C2 ·C2 + x3 C3 ·C2 = x2 kC2 k2 = 14x2 ,
i.e., x1 = −1.
V ·C3 = 18 = x1 C1 ·C3 + x2 C2 ·C3 + x3 C3 ·C3 = x3 kC3 k2 = 6x3 ,
i.e., x1 = 3. Hence the solution is


2
X = −1 .
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