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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 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 ) · u = (x1 − 1, x2 − 2, x3 − 3, x4 − 4) · (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 )·n = 0. This means
0 =(X − P )·(3, −6, 5) = (x1 − 1, x2 + 3, x3 + 4)·(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 × 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
5
−1
1
,
0
0
−1
1
,
0
0
0
is a basis of the
0
2 3
vector space of 2×2 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×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 · 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 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 ·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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