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Math 601 Solutions to Homework 4
- Use row reduction to compute the determinants of the following ma-
trices:
(a) A =
(b) B =
a a a a a a
b 2 b b b b b
c c 3 c c c c
d d d 4 d d d
e e e e 5 e e
Answer:
(a) We row reduce the matrix:
− row 1
−2 row 1
−4 row 1
×
(^) + row 2
+2 row 2
− row 3
The determinant of the row reduced matrix is 3 (because the ma-
trix is triangular, so the determinant is the product of the di-
agonal entries). Most of the row operations did not change the
determinant. When we multiplied the second row by
1 2
, we also
multiplied the determinant by
1 2
. This is the only row operation
that changed the determinant. Thus, the original matrix, had
determinant 2 · 3 = 6. Thus, det(A) = 6
(b) We can begin by dividing the first row by a, the second row by b,
the third row by c, the fourth row by d, and the fifth row by e.
We get:
a a a a a a
b 2 b b b b b
c c 3 c c c c
d d d 4 d d d
e e e e 5 e e
= abcde
Now, we can subtract the first row from each of the other rows:
The determinant of the resulting matrix is 2 · 3 · 4 · 5 = 120. Thus,
det(B) = 120abcde
- For what values of a is the following matrix invertible:
A =
a 0 4 − 3
4 2 a 1
1 0 1 − 1
Answer: We will use cofactor expansion to compute the determinant
of the matrix. We will expand along the second column:
det(A) = 1
a 4 − 3
4 a 1
a 4 − 3
a
a 1
4 a
a 4
= a(−a − 1) − 4(− 4 − 1) − 3(4 − a) + 2(−4 + 3) − 8(a − 4)
= −a
2 − 6 a + 38
The matrix A is invertible if det(A) 6 = 0, so we set the determinant
equal to zero and solve for a. If −a
2 − 6 a + 38 = 0, then a = − 3 ±
Thus, A is invertible for a 6 = − 3 ±
- Consider the subspace S of C[0, 1] spanned by cosh x and sinh x, and
consider the following two bases of the subspace S:
E = [cosh x, sinh x]
F =
[
e
x , e
−x
]
(a) Find the transition matrix from E to F.
(b) Let D be the differentiation operator on S; that is, D (f (x)) = f
′ (x).
Find the matrix representing D with respect to the basis E.
(c) Find the matrix representing D with respect to the basis F.
Answer:
(a) We write each of the vectors in E as a linear combination of the
vectors in F :
cosh x =
e
x
e
−x
sinh x =
e
x −
e
−x
Thus, the transition matrix from E to F is
(b) To find the matrix representing D with respect to the basis E, we
apply D to each of the basis vectors:
D(cosh x) = sinh x
D(sinh x) = cosh x
Thus, the matrix representing D with respect to the basis E is
[
]
(c) To find the matrix representing D with respect to the basis F , we
apply D to each of the basis vectors:
D(e
x ) = e
x
D(e
−x ) = −e
−x
Thus, the matrix representing D with respect to the basis F is
[
]
- Consider the following linear transformation from P 3 to P 2 :
L(p) = p
′′
′
Find the matrix representation of L with respect to the bases {1 + x, 1 − x, 1 + x
2 }
and {2 + x, x}.
Answer: We apply L to each of the vectors in the first basis:
L(1 + x) = 2
L(1 − x) = − 2
L(1 + x
2 ) = 2 + 4x
Now, we write each of the above polynomials as a linear combination
of the polynomials 2 + x and x. We get:
2 = 1(2 + x) − 1(x)
− 2 = −1(2 + x) + 1(x)
2 + 4x = 1(2 + x) + 3(x)
Thus, the matrix representing L with respect to the given bases is:
[
]
This is equivalent to:
− 6 a + 10b = 0
− 3 a + 5b = 0
Solving for a and b, we get that a = −
5 3
b. There are infinitely
many solutions. We are just asked for one solution, so we can
choose b = 3. Thus, one nonzero vector is
v =
[
]
Note: Any multiple of this vector also works.
(c) We apply A to u and v:
[
] [
]
[
]
[
] [
]
[
]
Now, we write these as linear combinations of u and v. We get:
[
]
= 2u + 0v
[
]
= 0u + 3v
Thus, the matrix representing A with respect to the basis [u, v] is
[
]
Note 1: This answer does not depend on which vectors we chose
for parts (a) and (b).
Note 2: Since we knew that Au = 2u and Av = 3v, we did not
need to do any computations to obtain this matrix.
