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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
Question 1: Find the equation of the hyperplane in R4 that passes through P = (1, 2, 3, 4)
and is orthogonal to u := (4, 3, 1, 1).
The hyperplane in question is the collection of points X = (x1 , x2 , x3 , x4 ) in R4 so that X − P is
orthogonal to u. This means that
0 = (X − P ) &middot; u = (x1 − 1, x2 − 2, x3 − 3, x4 − 4) &middot; (4, 3, 1, 1) = 4x1 + 3x2 + x3 + x4 − 4 − 6 − 3 − 4
The equation of hyperplane is
4x1 + 3x2 + x3 + x4 − 20 = 0
Question 2: Find an equation of the hyperplane in R3 that passes through P = (1, −3, −4)
and is parallel to the hyperplane H 0 determined by the equation 3x1 − 6x2 + 5x3 = 2.
The definition of H 0 implies that n = (3, −6, 5) is normal to H 0 . Since H and H 0 are parallel, n =
(3, −6, 5) is also a normal vector to H. By definition H is the collection of points X = (x1 , x2 , x3 )
in R3 so that (X − P )&middot;n = 0. This means
0 =(X − P )&middot;(3, −6, 5) = (x1 − 1, x2 + 3, x3 + 4)&middot;(3, −6, 5) =
= 3x1 − 3 − 6x2 − 18 + 5x3 + 20) = 3x1 − 6x2 + 5x3 − 1.
The equation of the hyperplane in question is
3x1 − 6x2 + 5x3 = 1.
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Mid term exam 1, October 11, 2011
Question 3: Let A =
1+i
4 + 2i
−2 − i
5
3−i
1
, B =
−6
2
0
−1
−1
. Find (a) A + B (b)
2
T
2AT − 3B .
2 − i −2 + i 2 + i
6 − 2i
4
−4
 
 
 
4 + 2i
1
2
2 + 2i − 3 8 + 4i − 6
−1 + 2i
10 + 3  = −4 − 2i
5 −3  0 −1 =  −4 − 2i
−6
−1 2
6 − 2i + 3 −12 − 6
9 − 2i
A+B =

1+i
T
2AT −3B = 2 −2 − i
3−i
Question 4: If possible find the inverses of A =
5
3
4
2
and B =
2
4

2 + 4i
13 
−18
3
.
6
The determinant of A is |A| = 10 − 12 = −2 =
6 0. The matrix is invertible. We use the formula
from class to compute the inverse:
1
−1 2
2 −4
−1
A =
= 3
− 52
−2 −3 5
2
The determinant of B is zero. B is not invertible.
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

1 2 3 4 5
0 0 1 2 5

Question 5: Consider the matrix A = 
0 0 0 1 5. What is the rank of A (explain)?
0 0 0 1 5
What is the dimension of the null space of A, N (A)? (explain).
Clearly

1
0
A∼
0
0
2
0
0
0
3
1
0
0
4
2
1
0

5
5
.
5
0
There are three pivots. This means that the rank of A is 3. There are two free variable; this means
that dim(N (A)) = 2
Question 6: Find 2 &times; 2 nonzero matrices A and B such that AB 6= BA.
Consider the following matrices:
A=
Then
0
0
1
AB =
0
1
,
0
B=
0
0
6=
0
0
0
1
0
,
0
0
= BA.
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Mid term exam 1, October 11, 2011

2
Question 7: Let A = 6
4
2
10
−4

3
7 . Find the LU factorization of A.
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We compute the echelon form of A to get U and L

 
2 2
3
2 2
A = 6 10 7  ∼ 0 4
4 −4 15
0 −8
 
1
l1 = 3 ,
2
Finally we have the following factorization

1 0
L = 3 1
2 −2
 
3
2 2
−2 ∼ 0 4
9
0 0


0
l2 =  1  ,
−2

3
−2
5
 
0
l3 = 0
1
A = LU where


0
2
0 , U = 0
1
0
2
4
0

3
−2
5
Question 8: Show that S = {1, 1 − X, (1 − X)2 } is a basis of the vector space P2 (X) (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 ({1, X, X 2 } is a basis). As a result
{1, 1 − X, (1 − X)2 } is also a basis since this linearly independent set contains 3 elements.
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1
1
1
1
,
1
1
5
−1
1
,
0
0
−1
1
,
0
0
0
is a basis of the
0
2 3
vector space of 2&times;2 matrices with real coefficients. Find the coordinate vector of A =
4 −7
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
2
4
3
x + x2 + x3 + x4
= 1
−7
x1 + x2
x1 − x2 − x3
x1
which in turn implies that
x1 = −7,
x1 + x2 = 4,
x1 − x2 − x3 = 3,
x1 + x2 + x3 + x4 = 2.
We then have x1 = −7, x2 = 11, x3 = −21, x4 = 19; as result, the coordinate vector of A relative
to S is
[A]S = (−7, 11, −21, 19)
Question 10: Let U be a n&times;n matrix with coefficients in R and assume that the columns of
U are nonzero and orthogonal in Rn . Prove that U is invertible.
Let us prove that the null space of U is zero. 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 &middot; C1 + x2 C2 &middot; C1 + . . . + xn Cn &middot; C1 = 0.
Since the columns of U are orthogonal C2 &middot; C1 = 0, . . . Cn &middot; 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 11, 2011
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 &middot;Cj for i 6= j, i, j ∈ {1, 2, 3}.
C1 &middot;C2 = −6 + 2 + 4 = 0,
C1 &middot;C3 = 2 + 2 − 4 = 0,
C2 &middot;C3 = −3 + 4 − 1 = 0.
This proves the statement.
(b) Consider the 3&times;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 &middot;C1 = 42 = x1 C1 &middot;C1 + x2 C2 &middot;C1 + x3 C3 &middot;C1 = x1 kC1 k2 = 21x1 ,
i.e., x1 = 2. Similarly
V &middot;C2 = −14 = x1 C1 &middot;C2 + x2 C2 &middot;C2 + x3 C3 &middot;C2 = x2 kC2 k2 = 14x2 ,
i.e., x1 = −1.
V &middot;C3 = 18 = x1 C1 &middot;C3 + x2 C2 &middot;C3 + x3 C3 &middot;C3 = x3 kC3 k2 = 6x3 ,
i.e., x1 = 3. Hence the solution is


2
X = −1 .
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