Winter 2017 Ma 1b Analytical Problem Set 4 Solutions
1. (5 pts) From Ch. 2.20 in Apostol: Problem 9
Solution. We are given the system of simultaneous equations:
2. (5 pts) From Ch. 2.20 in Apostol: Problem 10
Solution.
x + y + 2z = 2
2x − y + 3z = 2
5x − y + az = 6
(a) We are given the system of simultaneous equations:
5x + 2y − 6z + 2u = −1
We turn this into augmented matrices and solve
x − y + z − u = −2

1
 2
5

1
∼ 0
0

1
∼ 0
0

1 2 2
−1 3 2  R2 = R2 − 2R1
R3 = R3 − 5R1
−1 a 6

1
2
2
−3
−1
−2 
R3 = R3 − 2R2
−6 a − 10 −4

1
2
2
−3 −1 −2 
0 a−8 0
When a 6=
yielding

1
 0
0

1

∼ 0
0

1

∼ 0
0
8, we can divide the last row by a − 8

2
1
2
R1 = R1 − 2R3
−3 −1 −2 
R2 = R2 + R3
0
1
0

1 0 2
R1 = R1 + 13 R3
−3 0 −2 
R2 = − 31 R2
0 1 0

0 0 43
1 0 23 
0 1 0
So, we have a unique solution (x, y, z) = ( 43 , 23 , 0).
If on the other hand, a = 8, we get


1 1
2
2
 0 −3 −1 −2 
0 0
0
0
Thus, setting z = t, and reading off each row gives
−3y − t = −2 and hence, y = 23 − 31 t. The other
equation is x + 53 t = 34 yielding x = 34 − 53 t. So, our
solution set is
4 2
5 1
(x, y, z) =
, ,0 + t − ,− ,1 .
3 3
3 3
We turn this into augmented matrices and
solve
∼
5 2 −6 2 −1
1 −1 1 −1 −2
1 −1 1 −1 −2
5 2 −6 2 −1
swap rows
1 −1
1
−1 −2
0 7 −11 7
9
4
5
1 0 −7 0 −7
∼
0 1 − 11
1 97
7
R2 = R2 − 5R1
∼
R1 = R1 + 71 R2
R2 = 17 R2
Thus, setting z = s and u = t, it follows from
the rows that x = − 75 + 47 s and y = 97 + 11
7 s−t
giving us
5 9
(x, y, z, u) = − , , 0, 0
7 7
4 11
+s
, , 1, 0
7 7
+ t (0, −1, 0, 1)
(b) Since we are adding the equation
x+y+z =6
to the previous system of simultaneous equations, we can simply add it to our reduced augmented matrix:
1
Winter 2017 Ma 1b Analytical Problem Set 4 Solutions
And thus,
1 0 − 47 0 − 75
 0 1 − 11 1 9 
7
7
1 1
1
0 6
R3 = R3 − R1


1 0 − 47 0 − 75
1 97 
∼  0 1 − 11
7
0 1 11
0 47
R3 = R3 − R2
7
7


4
5
0 −7
1 0 −7
9 
1
∼  0 1 − 11
7
7
38
7
0 0 22
−1
R3 = 22
R3
7
7


5
4
−7
1 0 −7
0
4
11
9  R1 = R1 + 7 R3

1
∼ 0 1 −7
7
R2 = R2 + 11
19
7
7 R3
0 0
1
− 22
11


3
2
1 0 0 − 11
11
1
4 
∼ 0 1 0
2
7
19
0 0 1 − 22 11


Thus, setting u = t, it follows from the rows
3
2
that x = 11
+ 11
t and y = 4 − 21 t and
7
z = 19
11 + 22 t giving us
3
19
2
1 7
(x, y, z, u) =
, 4, , 0 +t
,− , ,1
11
11
11 2 22
3. (5 pts) From Ch. 2.20 in Apostol: Problem 13
Solution. We shall find the inverse of the matrix
A by augmenting [A|I] to [I|A−1 ].


1 2 2 1 0 0
 2 −1 1 0 1 0  R2 = R2 − 2R1
1 3 2 0 0 1
R3 = R3 − R1


1 0 0
1 2
2
∼  0 −5 −3 −2 1 0 
R2 ↔ R3
0 1
0 −1 0 1


1 2
2
1 0 0
R1 = R1 − 2R2
0 −1 0 1 
∼ 0 1
0 −5 −3 −2 1 0
R3 = R3 + 5R2


1 0 2
3 0 −2
R1 = R1 + 23 R3
∼  0 1 0 −1 0 1 
0 0 −3 −7 1 5
R3 = − 13 R3


2
4
1 0 0 − 53
3
3
1 
∼  0 1 0 −1 0
7
1
0 0 1 3 − 3 − 53
−1  5
2
−3
1 2 2
3
 2 −1 1  =  −1 0
7
1 3 2
− 13
3

4
3

1 
− 53
4. (5 pts) From Ch. 3.6 in Apostol: Problem 2
Solution. Note, that adding any scalar multiple
of one row of a square matrix to any other row does
not change the value of the determinant. On the
other hand, multiplying a row by a scalar changes
the determinant by that same multiple.
We are given the matrix


x y z
A =  3 0 2 .
1 1 1
We are also given that |A| = 1.
(a) We get our new matrix by multiplying the first
row of A by 2 and the second row of A by 12 .
Thus,
2x 2y 2z 3
= 2 · 1 · 1 = 1.
0
1
2
2
1 1 1 (b) We get our new matrix by adding three times
the first row of A to the third row of A and
one times the first row of A to the third row
of A, so the determinant does not change. ie.
x
y
z
3x + 3
= 1.
3y
3z
+
2
x+1 y+1 z+1 (c) We get our new matrix by adding −1 times
the third row of A to the first row of A and 1
times the third row of A to the second row of
A, so the determinant does not change. ie.
x−1 y−1 z−1 4
= 1.
1
3
1
1
1 2
Winter 2017 Ma 1b Analytical Problem Set 4 Solutions
5. (5 pts) From Ch. 3.6 in Apostol: Problem 3
and
1 1 1 2 2 2 a b c a3 b3 c3 Solution.
(a)
1 1 1
a b c
a2 b2 c2
(R2 = R2 − a2 R1, R3 = R3 − a3 R1)
1
1
1
2
2
2
2
= 0 b − a c − a 0 b3 − a3 c3 − a3 (R2 = R2 − aR1, R3 = R3 − a2 R1)
1
1
1
c − a = 0 b − a
0 b2 − a2 c2 − a2 (R3 = R3 − (b + a)R2)
1
1
1
c−a
= 0 b − a
0
0
c2 − bc − ac + ba
(R3 = (b + a)R3 − (b2 + ab + a2 )R2)
1
1
1
1 2
2
2
2
0 b − a c − a =
b+a
0
0
(∗) ∗ = (b + a)(c3 − a3 ) − (b2 + ab + a2 )(c2 − a2 )
1
· 1 · (b2 − a2 ) · (∗)
=
b+a
=(b − a) · (∗)
=1 · (b − a) · (c2 − bc − ac + ba)
To get a nicer form for this we analyse (∗).
Note that when c = a we clearly have (∗) =
0. Moreover, when c = b, we have (∗) =
(c + a)(c3 − a3 ) − (c2 + ac + a2 )(c2 − a2 ) =
(c4 + ac3 − a3 c − a4 ) − (c4 + ac3 − a3 c − a4 ) = 0.
So (∗) has the factors (c − a) and (c − b). Polynomial division gives
=(b − a)(c − a)(c − b)
Note, that we have used that the determinant
of an upper triangular matrix is simply the
product of the diagonal.
(b)
1 1 1
a b c
a3 b3 c3
(∗) = (c − a)(c − b)(bc + ac + ab)
And
1
2
a
a3
(R2 = R2 − aR1, R3 = R3 − a3 R1)
1
1
1
c − a = 0 b − a
0 b3 − a3 c3 − a3 (R3 = R3 − (b2 + ab + a2 )R2)
1
1
1
c−a
= 0 b − a
0
0
c3 − b2 c − a2 c + b2 a + a2 b − abc
thus, the determinant is
1 1 b2 c2 = (b−a)(c−b)(c−a)(ab+ac+bc).
b3 c3 Note, if you’re worried about the factorisation
of (∗), here is an example (let c − a = α):
∗ = (b + a)(c3 − a3 ) − (b2 + ab + a2 )(c2 − a2 )
= α (b + a)(c2 + ac + a2 ) − (b2 + ab + a2 )(c + a)
=1 · (b − a) · (c3 − b2 c − a2 c + b2 a + a2 b − abc)
= α(bc2 + ac2 − b2 c − ab2 ) − expanding
=(b − a)(c − a)(c2 + ac − b2 − ab)
= α(c − b) (bc + a(c + b))
=(b − a)(c − a)(c − b)(a + b + c)
= (c − a)(c − b)(ab + ac + bc)
3