Download Math 601 Solutions to Homework 3

Math 601 Solutions to Homework 3
1. Use Cramer’s Rule to solve the following system of linear equations.
(Solve for x1 , x2 , and x3 in terms of a, b, and c.)
2x1 − x2 + 7x3 = a
5x1 − 2x2 − x3 = b
3x1 − x2 − 3x3 = c
Answer: Let A be the matrix corresponding to the left side of the
system of linear equations,
 andlet Ai denote the matrix A with column
a

b . Then, to use Cramer’s Rule, we must
i replaced by the vector
c
compute the following determinants:
¯
¯
¯ 2 −1
¯
7
¯
¯
¯
det(A) = ¯ 5 −2 −1 ¯¯ = 5
¯ 3 −1 −3 ¯
¯
¯ a −1
7
¯
det(A1 ) ¯¯ b −2 −1
¯ c −1 −3
¯
¯ 2 a
7
¯
¯
det(A2 ) = ¯ 5 b −1
¯ 3 c −3
¯
¯
¯
¯ = 5a − 10b + 15c
¯
¯
¯
¯
¯
¯ = 12a − 27b + 37c
¯
¯
¯
¯
¯ 2 −1 a ¯
¯
¯
det(A3 ) = ¯¯ 5 −2 b ¯¯ = a − b + c
¯ 3 −1 c ¯
Now, we can use Cramer’s rule to find the solution:
det(A1 )
5a − 10b + 15c
x1 =
=
= a − 2b + 3c
det(A)
5
x2 =
12a − 27b + 37c
12
27
37
det(A2 )
=
= a− b+ c
det(A)
5
5
5
5
x3 =
det(A3 )
a−b+c
1
1
1
=
= a− b+ c
det(A)
5
5
5
5
1
2. Determine whether the following sets are subspaces of R2×2 . Explain
your answers.
¯
©
ª
(a) S1 = A ∈ R2×2 ¯ a12 − a22 = 0
¯
ª
©
(b) S2 = A ∈ R2×2 ¯ a11 + a22 = 1
½
¯ µ
2×2 ¯
(c) S3 = A ∈ R
¯
2 1
−3 4
¶
µ
A=A
2 1
−3 4
¶¾
¯
©
ª
(d) S4 = A ∈ R2×2 ¯ det(A) = 0
Answer:
(a) Yes, S1 is a subspace of R 2×2 . To show that S1 is a subspace, we
must show that it is closed under scalar multiplication and closed
under addition.
(1) Closed under scalar multiplication:
Suppose that A is a matrix in S1 . Then,
a12 − ¶
a22 = 0, so
µ
a11 a12
a12 = a22 and we can write A as A =
.
a21 a12
µ
¶
αa11 αa12
Consider αA =
. The 1, 2 and 2, 2 entries
αa21 αa12
of αA are equal, so αA ∈ S1 , and S1 is closed under scalar
multiplication.
(2) Closed under addition:
Suppose that A and B are both matrices in S1 . Then,
µ a12 = a¶22
a11 a12
and b12 = b22 , so we can write A and B as A =
a21 a12
¶
µ
b11 b12
.
and B =
b21 b12
¶
µ
a11 + b11 a12 + b12
. The 1, 2 and 2, 2
Consider A + B =
a21 + b21 a12 + b12
entries of A + B are equal, so A + B ∈ S1 .
2
We have shown that S1 is closed under addition and scalar multiplication. Thus, S1 is a subspace of R 2×2 .
(b) No, S2 is not a subspace of R 2×2 . To show that S2 is not a
subspace, we will show that it is not closed under scalar multiplication.
µ
¶
1 0
Consider the matrix A =
, which is in S2 .
0 0
µ
¶
2 0
If we multiply A by the scalar 2, we get 2A =
, which is
0 0
not in S2 .
Thus, S2 is not a subspace of R 2×2 .
(c) Yes, S3 is a subspace of R 2×2 . To show that S3 is a subspace, we
must show that S3 is closed under scalar multiplication and closed
under addition.
(1) Closed under scalar multiplication:
µ
¶
2 1
Let M =
.
−3 4
Suppose that A is a matrix in S3 . Then, M A = AM . Consider αA. We would like to show that M (αA) = (αA)M , but
this just follows from properties of matrix multiplication and
the assumption that M A = AM :
M (αA) = α(M A) = α(AM ) = (αA)M
Thus, αA ∈ S3 . Thus, S3 is closed under scalar multiplication.
(2) Closed under addition:
Suppose that A and B are matrices in S3 . Then, M A = AM
and M B = BM . Consider A + B. We would like to show
that M (A+B) = (A+B)M . This follows from the properties
of matrix multiplication and the assumption that M A = AM
and M B = BM :
M (A + B) = M A + M B = AM + BM = (A + B)M
Thus, A + B ∈ S3 . Thus, S3 is closed under addition.
3
We have shown that S3 is closed under scalar multiplication and
closed under addition. Thus, S3 is a subspace of R 2×2 .
(d) No, S4 is not a subspace of R 2×2 . To show that S4 is not a
subspace, we will show that S4 is not closed under addition.
µ
Consider the matrices A =
1 1
0 0
¶
µ
and B =
0 0
1 2
¶
.
Since det(A) = 0 and det(B)
µ = 0,¶both A and B are in the set S4 .
1 1
Now, consider A + B =
. Since, det(A + B) = −1 6= 0,
1 2
the matrix A + B is not in the set S4 .
Thus, S4 is not closed under addition, so it is not a subspace.
3. Let P4 denote the vector space of all polynomials with degree less than
or equal to 3. Consider the subspace of P4 consisting of polynomials
p(x) for which p(−1) = 0. Find a basis for this subspace.
¯
n
o
¯
Answer: We are considering the subspace S = p(x) ∈ P4 ¯ p(−1) = 0 .
A polynomial in P4 is of the form p(x) = a + bx + cx2 + dx3 . If the
polynomial is in the the subspace S, then p(−1) = 0, which means
(plugging x = −1 into the polynomial p(x)) that a − b + c − d = 0. If
we solve this equation for a, we get a = b − c + d.
Thus, every polynomial in S can be written in the form
p(x) = (b − c + d) + bx + cx2 + dx3 . We can rewrite this as
p(x) = b(1 + x) + c(−1 + x2 ) + d(1 + x3 )
Thus, every polynomial in S can be written as a linear combination of
the polynomials 1 + x, −1 + x2 , and 1 + x3 . This means that
Span(1 + x, −1 + x2 , 1 + x3 ) = S
4
All we need to do now is show that these three polynomials are linearly
independent:
Suppose there exists c1 , c2 , c3 such that
c1 (1 + x) + c2 (−1 + x2 ) + c3 (1 + x3 ) = 0
Rearranging, we get
(c1 − c2 + c3 ) + c1 x + c2 x2 + c3 x3 = 0
Since this must be true for all values of x, the coefficients on the left
side must all be 0. Thus, using the coefficients of the x, x2 , and x3
terms, we get that c1 = 0, c2 = 0, and c3 = 0.
Thus, the polynomials 1+x, −1+x2 , and 1+x3 are linearly independent.
We have shown that the polynomials 1 + x, −1 + x2 , and 1 + x3 span
S and are linearly independent. Thus, they form a basis for S.
A basis for the subspace S is {1 + x, −1 + x2 , 1 + x3 }.
4. Let C[−π, π] denote the vector space of all real-valued functions that
are defined and continuous on the closed interval [−π, π].
(a) Consider the subspace of C[−π, π] spanned by the vectors cos x,
sin x, and sin(2x). What is the dimension of this subspace? Explain your answer.
(b) Consider the subspace of C[−π, π] spanned by the vectors cos x,
cos(3x), and cos3 x. What is the dimension of this subspace?
Explain your answer.
Answer:
(a) The main thing we need to do is determine whether the vectors
are linearly independent.
Suppose there exists c1 , c2 , c3 such that
5
c1 cos x + c2 sin x + c3 sin(2x) = 0
This must hold for all values of x, so we can plug in specific values
of x and get linear equations involving c1 , c2 , and c3 . We will plug
π
π
in x = 0, x = , and x = :
2
4
x=0
c1 +
0+ 0=0
x=
π
2
0+
c2 + 0 = 0
x=
π
4
1
1
√ c1 + √ c2 + c3 = 0
2
2
Solving this system of linear equations, we get c1 = 0, c2 = 0,
and c3 = 0. Thus, the vectors cos x, sin x, and sin(2x) are linearly
independent. Since these vectors also span the subspace, they
form a basis for the subspace. Thus, the dimension of the subspace
is 3.
(b) Again, we need to determine if the vectors are linearly independent. In this case, the vectors are linearly dependent due to the
following trig identity:
cos(3x) = −3 cos x + 4 cos3 x
Here is one way to see that this trig identity holds:
cos(3x) =
=
=
=
=
=
=
Re(ei3x )
Re((cos x + i sin x)3 )
Re(cos3 x + 3i cos2 x sin x − 3 cos x sin2 x − i sin3 x)
cos3 x − 3 cos x sin2 x
cos3 x − 3 cos x(1 − cos2 x)
cos3 x − 3 cos x + 3 cos3 x
−3 cos x + 4 cos3 x
6
Next, we choose 2 of the vectors and see if they are linearly independent
(it does not matter which 2 we choose). We will choose cos x and cos 3x.
Suppose there exists c1 and c2 such that
c1 cos x + c2 cos(3x) = 0
Since this must hold for all values of x, we can plug in specific values
of x and get linear equations involving c1 and c2 . We will plug in x = 0
π
and x = . This gives us the equations
3
c1 + c2 = 0
1
c1 − c3 = 0
2
Solving, we get c1 = 0 and c2 = 0. Thus, the two vectors are linearly
independent. They also span the subspace, and so they form a basis
for the subspace. Thus, the dimension of the subspace is 2.
7