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Homework 2 Solutions
Course Reader Exercises
5.9 The goal is to find scalars a and b such that
a
[
]
[
]
[
]
This leads to the system of equations
2 a + 3 b = 1
1 a + 2 b = 2
The second equation implies a = 2 − 2 b, so inserting this into the first equation gives
4 − 4 b + 3b = 1, or b = 3. Plugging back into either equation gives a = −4, so
[
]
[
]
[
]
5.10 We want to know if there exist scalars c 1 , c 2 , c 3 such that
c 1
(^) + c 2
(^) + c 3
This leads to the system of equations
2 c 1 + 5 c 2 + c 3 = 8
c 1 + 8 c 2 + 6 c 3 = − 5
c 1 − 2 c 3 = − 11
which has augmented matrix
A =
Since
rref(A) =
the last equation of the reduced system is 0 = 1, so there are no solutions. Thus v is
not in the span of the given set of vectors.
5.11 We need to find a solution to the system of equations that corresponds to the following
matrix of coefficients, which we row reduce :
So we conclude that
5.12 We want to solve
c 1
The augmented matrix corresponding to this system is
A =
Since
rref(A) =
the reduced system is
c 1 + 5 c 3 = − 3
c 2 − c 3 = 4
Choosing c 3 = 0 we get c 1 = −3 and c 2 = 4. Thus
There are infinitely many solutions — every choice of c 3 yields a different linear com-
bination which equals v.
(b) Again we switch rows 1 and 2, and the reduced row echelon form is
The last row says 0 = −2, so the system is inconsistent. No solutions.
6.13 From the corresponding equations p(1) = 1, p(2) = 2, and p(−1) = 5 we get the
equations
a + b + c = 1
4 a + 2 b + c = 2
a − b + c = 5
The augmented matrix for this system is
with reduced row echelon form
So the solutions are c = 2, b = −2, and a = 1, and the polynomial is p(x) = x
2 − 2 x+2.
6.14 Substituting the given information, we obtain the following system
− 8 + 4 a − 2 b + c = 2
− 1 + a − b + c = 3
1 + a + b + c = 0
8 + 4 a + 2 b + c = 8
Moving the constant terms to the right hand side, we obtain a system whose augmented
matrix is (^)
The reduced row echelon form is
Thus the unique solution is a = 7/6, b = − 5 /2, c = 1/3, which means
p(x) = x
3
x
2 −
x +
satisfies all of the given conditions. (Check!)
6.15 We have
f (t) = A cos t + B sin t + Ce
t
f
′ (t) = −A sin t + B cos t + Ce
t
f
′′ (t) = −A cos t − B sin t + Ce
t
so the three conditions f (0) = 2, f
′ (0) = 0 and f
′′ (0) = 6 lead to the system
A + C = 2
B + C = 0
−A + C = 6
The reduced row echelon form of the system is
A = − 2
B = − 4
C = 4
so A = −2, B = −4 and C = 4.
8.10 The observation to make here is that x = c is a solution of Ax = Ac. By Proposition
11.1, every other solution takes the form c + xn where xn ∈ N (A). The null space was
found in Exercise 64.
(a) x =
(b) x =
(c) x =
π/ 6 √ e 3
8.12 Since
rref(A) =
[
]
9.2 Performing elimination on the augmented matrix
b 1
b 2
b 3
we arrive at (^)
b 1
b 1
b 2
5 b 1
b 2
− b 3
Although this is not the reduced row echelon form, we need not continue further, since
we can already see where the inconsistencies could possibly arise. From the third row,
we see that the components of b must satisfy
5 b 1
b 2
− b 3 = 0
to avoid inconsistency.
9.3 Performing elimination on the augmented matrix
b 1
b 2
b 3
b 4
we arrive at (^)
b 1 1 3
b 3
b 2 + 2b 1 − b 3
b 4 − 3 b 1 + b 3
Although this is not the reduced row echelon form, we need not continue further, since
we can already see where the inconsistencies could possibly arise. From the third and
fourth rows, we see that the components of b must satisfy
2 b 1 + b 2 − b 3 = 0
− 2 b 1 b 3 + b 4 = 0
to avoid inconsistency.
9.4 (b) Since 2(0) + 3 − 3 = 0 and −2(0) + 3 + (−3) = 0,
is in the column space
of A.
(c) Since 2(1) + 2 − 4 = 0 and −2(1) + 4 + (−2) = 0,
is in the column space
of A.
(e) Since −2(2) + 3 + (−1) = − 2 6 = 0,
is not in the column space of A.
