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Unformatted text preview: B U Department of Mathematics
Math 201 Matrix Theory
Summer 2005 First Midterm
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1.) a)[4] Find c such that the following set of columns is a basis for R : 1
2
1
, 1 , 1 .
1 −1
0
c Solution: 1 21
12
1
12
1
−r1 +r2 →r
2r +r →r
0 − 2− 3 −3 0 −1
0 . Hence
A = 1 1 1 − − − −2 0 −1
− − −→
−− →
−
r1 +r3 →r3
−1 0 c
0 2 c+1
0 0 c+1
c = −1, i.e., ∀c ∈ R \ {−1} the given set of columns is a basis for R3 .
b)[4] Is the set of polynomials S = {1 − x, 1 + x, 1 − x2 } linearly independent?
Solution:
Consider
a(1 − x) + b(1 + x) + c(1 − x2 ) = 0
Then −cx2 = 0 implies c = 0. So a + b = 0 and −a + b = 0 give that a = 0, b = 0. Thus
S is linearly independent.
c)[2] If a matrix A is n × (n − 1) and its rank is (n − 2) what is the dimension of its null space?
Solution: 2.) Since the dimension of the null space is the diﬀerence of the number of unknowns and
the rank, we get
dim(Null(A)) = (n − 1) − (n − 2) = 1 1
21
Let A = 2 −1 1 −1 3 0 a) [10] Find the LU decomposition of A.
Solution: 1
21
12
1
12
1
−2r1 +r2 →r
r2 +r →r3
− − −→
−−−
A = 2 −1 1 − − − −2 0 −5 −1 − − 3− → 0 −5 −1 = U , where
r1 +r3 →r3
−1 3 0
05
1
00
0
−1 −1 −1
E3 E2 E1 A = U , i.e., A = E1 E2 E3 U . Writing explicitly 100
1 00 1 0 0 1 2
1
A = 2 1 0 0 1 0 0 1 0 0 −5 −1
0 0 1 −1 0 1 0 −1 1 0 0
0 1
1
0012
1 0 0 −5 −1
= 2
−1 −1 1 0 0
0 1
00
2
1 0
where L =
−1 −1 1 b) [6] Find a basis for the column space and the null space of A. What is the rank of A?
Solution:
From U we see pivots 1 and 5 appear in the ﬁrst and second columns. Therefore that
2
1 2 , −1 is a basis for the column space of A. To ﬁnd a basis for the null space −1
3
recall that Ax = 0 ⇐⇒ U x = 0. Then
x1 + 2x2 + x3 = 0
−5x2 − x3 = 0 3
3 1 . Thus 1 is a basis for the
implies x3 = −5x2 and x1 = 3x2 , hence, x = x2 −5
−5
null space of A. Since rank equals to the dimension of the column space, rank (A) = 2.
c) [4] Using the LU decomposition of A ﬁnd the complete solution to 4
3
Ax =
1
Solution: 4
4 3 implies Ly = 3 , since Ax = LU x. Then using L
Setting y = U x, Ax =
1
1
from part (a),
y1
=4
2y1 +y2
=3
−y1 −y2 +y3 = 1 4
entails y1 = 4, y2 = −5 and y3 = 0. Now, U x = −5 gives that
0
x1 +2x2
x3
=4
−5x2 −x3 = −5 −1
3 0 + x2 1 .
hence x1 = 3x2 − 1 and x3 = −5x1 + 5, i.e., x =
5
−5
3.) a) [6] Let A be an m × n and B be an n × m matrix, and m > n. What can you say about the
invertibility of AB ?
Solution:
We claim that AB is singular. Given m > n there exists a nonzero solution to Bx = 0,
i.e., ∃x0 = 0 such that Bx0 = 0. Then (AB )x0 = A(Bx0 ) = 0. But AB being an
m × m matrix and (AB )x0 being zero with x0 = 0 implies dim(Null(AB )) = 0, hence
rank (AB ) = m. Thus AB is not invertible. b) [6] Let A and B be n × n matrices. Show that if A is singular then AB is also singular.
Solution:
Assume that A is singular. Then AT is also singular, i.e., AT x has a nontrivial solution,
say, AT x0 = 0 for some x0 = 0. But then we get that (AB )T x0 = B T (AT x0 ) = 0, so that
(AB )T is singular. Thus AB is singular.
c) [3] If A is an n × n matrix with A2 = A and rank(A)=n, ﬁnd A.
Solution:
rank (A) = n implies that A is invertible, i.e., A−1 exists. Then multiplying both sides of
A2 = A by A−1 we get
A2 A−1 = AA−1 = I
and so
A = I.
4.) Let T : R3 → R3 be deﬁned by T (x, y, z ) = (x + 2y + z, x + y, 2y + z ).
a) [2] Write down what we must show to prove that T is a linear transformation. (Do not carry
out the computations).
b) [5] What is the matrix representing this transformation in the standard basis for R3 .
c) [8] Show that T is nonsingular and ﬁnd its inverse transformation.
Solution:
a) We have to show that given two points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) in R3
T ((x1 , y1 , z1 ) + (x2 , y2 , z2 )) = T (x1 , y1 , z1 ) + T (x2 , y2 , z2 ),
T (c(x1 , y1 , z1 )) = cT (x1 , y1 , z1 ), ∀c ∈ R.
b) Since T (1, 0, 0) = (1, 1, 0), T (0, 1, 0) = (2, 1, 2) and T (0, 0, 1) = (1, 0, 1), we get that
the matrix representing T is 121
A = 1 1 0 .
021
c) To show that T is nonsingular, it suﬃces to show that A is row equivalent to I3×3 .
Using GaussJordan method we get that 12110
[AI ] = 1 1 0 0 1
02100 1 2 0 −1
r2 −r →r2
− − 3− → 0 1 0 −1
−−−
r1 −r3 →r1
001 2 0
12
r −r →r2
0 −1− 2− → 0 1
−−−
2r2 −r3 →r3
1
00 2
1
r1 −2r →r
1
1 − − 2 −1 −− →
−
−2 −1 11 0
0
1 1 −1 0 1 2 −2 −1 100 1
0 −1
1 = [I A−1 ].
0 1 0 −1 1
0 0 1 2 −2 −1 . Knowing A−1 1
0 −1
1 , we get that
= −1 1
2 −2 −1
T −1 (x, y, z ) = (x − z, −x + y + z, 2x − 2y − z ). ...
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