10 Systems of Linear Equations
Sheet 10. Systems of Linear Equations
Exercise 10.1. Solve the following systems:
(a) by means of Cramer's Rule,
(b) by means of elementary operations.
a)
x1 − 3x2 + 5x3 = −4 2x1 + 5x2 − x3 = 3
− x1 − x2 + 3x3 = −4 b)
− x1 + 2x2 − x3 = 2 3x1 − x2 + x3 = 12 2x1 + 8x2 − 3x3 = 12
c)
5x1 − 3x2 + 7x3 = 0
− 4x1 + x2 − 5x3 = 0 x1 − x2 + x3 = 0
d)
x2 − 3x3 + 4x4 = 0
x1 − 2x3 = 0
3x1 + 2x2 − 5x4 = 2
4x1 − 5x3 = 0
e)
2x − y + z = 2 3x + 2y + 2z = −2 x − 2y + z = 1
f)
x + 2y − 3z = 0 4x + 8y − 7z + t = 1 x + 2y − z + t = 1
− x + y + 4z + 6t = 0
g)
1 1 −1 1 −3 2
−1 2 −1
x1 x2 x3
=
−2 0 1
h)
x1 − 2x2 = −2 2x2 + x3 = 1 x1 − x3 = 1
i)
x1 + 3x2 − x3 = 8 x1 + x2 − 3x3 = 2 2x2 + x3 = 5
j)
x1 + x2 + x3 = 0 2x1 − x2 − x3 = −3 x1 − − x2 + x3 = 0 Exercise 10.2. Solve the following systems:
a)
"
2 5 −7 3 −8 5
#
x1 x2
x3
=
"
1 2
#
b)
"
1 −2 5 1 2 −3
#
x1 x2
x3
=
"
0 0
#
c)
1 43 −3 3 2 −5 3 4 −9 5 2 −8
x1 x2 x3
=
3 0 9 0
d)
x1 − 2x2 = 2
− 2x1 + 4x2 + x3 = 3
− x1 + 2x2 + x3 = 1
Last update: January 7, 2009 1
10 Systems of Linear Equations
Exercise 10.3. Find the general solutions of the following systems. Indicate two distinct particular solutions:
a)
2x1 − x2 + x3 − x4 = 0
− x1 + 3x2 − x3 + 2x4 = 0 x1 + 2x2 + x4 = 0
b)
− x1 + 2x3 + x4 = 0 2x1 − x2 + 2x3 − x4 = 0 x1 + 2x2 − x3 = 0 2x1 + x2 + 3x3 = 0
c)
x1 + x2 + 2x3 + 3x4 = 0 2x1 − 3x2 + 4x3 + x4 = 0 4x1 − x2 + 8x3 + 7x4 = 0
d)
x1 − x2 + 2x3 = −3 x2 + x3 = −2 x1 − 2x2 + x3 = −1 Exercise 10.4. Find one of the basis solutions of the following systems:
a)
( x1 − x2 + x3 − x4 = −1
2x1 + x2 − 3x3 + x4 = 5 b)
x1 − 2x2 = 1 2x2 + x3 = −1 x1 + x3 = 0
c)
x1 − x2 + x3 + 2x4 = 0
− x1 + 2x2 + x3 = 1
− 2x2 + 3x4 = −2
d)
− 2x1 + x3 + x4 = 5
x1 + x2 − x3 = −2
3x1 + 2x3 + x5 = −2 Exercise 10.5. Find all basis solutions of the systems:
a)
( x1 − 2x2 + 3x4 = 2
− 2x1 + 4x2 + x3 + x4 = 3 b)
x1 + x2 + x3 = 1
− 2x1 + 2x2 + 3x3 = −1
− x1 + 3x2 + 4x3 = 0
c)
− 5x2 + x5 = 3
+ 6x2 + x4 = 3
x1 − 7x2 = 2
8x2 + x3 = 1
d)
x1 + x2 + x3 = 3
− 2x1 + 2x2 + 3x3 = 3
− x1 + 3x2 + 4x3 = 6
Exercise 10.6. Solve the system of equations by means of elementary operations on rows. What are the basis variables and the free variables of the found general solution. Indicate two distinct particular solutions of the system, one of them should be the basis solution. Find the conditions for the general solution to be nonnegative
a)
x + 2y + 3z + t = 1 2x + 4y − z + 2t = 2 3x + 6y + 10z + 3t = 3 x + y + z + t = 0
b)
2x + y + z + 2t = 16 x − 2y + 3z + t = 3
− x + y + 2z − 5t = −9
x + y + z = 8
Last update: January 7, 2009 2
10 Systems of Linear Equations
Exercise 10.7. Solve the system of equations
6x1 + 3x2 − x3 + x4 = 3 x2 + x3 − x4 = 1 3x1 + 3x2 + x3 − x4 = 3
with respect to the indicated variables, characterize the set of nonnegative solutions.
(a) The basis variables x1, x2, (b) the basis variables x3, x4,
(c) the basis variables x2, x3.
Answers:
Exercise 10.1.
a)
x1 = 1 x2 = 0 x3 = −1
b)
x1 = 1 x2 = 12 x3 = −2
c)
x1 = 0 x2 = 0 x3 = 0
d)
x1 = 0 x2 = 138 x3 = 0 x4 = −132
e)
x = 2 y = −1 z = −3
f)
t = 1 x = 4 y = −2 z = 0
g)
x1 = −1 x2 = 1 x3 = 2
h)
x1 = 0 x2 = 1 x3 = −1
i)
x1 = 3 x2 = 2 x3 = 1
j)
x1 = −1 x2 = 0 x3 = 1 Exercise 10.2.
a)
x1 = 1831+ x3 x2 = −311 + x3 x3 ∈ R
b)
x1 = −x3 x2 = 2x3 x3 ∈ R
c)
x1 = −277 x2 = −149 x3 = −187
d) No solution
Exercise 10.3.
a)
x1 = 15x4− 25x3 x2 = 15x3− 35x4 x3 ∈ R
x4 ∈ R
b)
x1 = 137 x4 x2 = −135 x4 x3 = −133 x4 x4 ∈ R
c)
x1 = −2x3− 2x4 x2 = −x4
x3 ∈ R x4 ∈ R
d)
x1 = −3x3− 5 x2 = −x3− 2 x3 ∈ R
Exercise 10.4.
a)
x1 = 23x3+43 x2 = 53x3− x4+73 x3 ∈ R
x4 ∈ R
b)
x1 = −x3 x2 = −12x3−12 x3 ∈ R
c)
x1 = 54x4+ 1 x2 = 53x3− x4+ 73 x3 ∈ R
x4 ∈ R
d)
x1 = 27x4− 17x5−127 x2 = 32x4+ 1
x3 = −74x4 x4 ∈ R x5 ∈ R
Last update: January 7, 2009 3
10 Systems of Linear Equations
Exercise 10.5.
a)
x1 = 2x2− 3x4+ 2 x2 ∈ R
x3 = 7 − 7x4 x4 ∈ R
b)
x1 = 14x3+ 34 x2 = 14 − 54x3 x3 ∈ R
c)
x1 = 75x5− 115 x2 = 15x5− 35 x3 = 295 − 85x5 x4 = 335 − 65x5 x5 ∈ R
d)
x1 = 14x3+34 x2 = 94 −54x3 x3 ∈ R
Exercise 10.6.
a)
x = t − 1 y = 1 z ∈ R z = 0
b)
x = 8 − t y = t + 1 z = t − 1 t ∈ R Exercise 10.7.
a)
x1 = 23x3− 23x4 x2 = x4− x3+ 1 x3 ∈ R
x4 ∈ R
b) impossible c)
x1 ∈ R x2 = 1 − 32x1 x3 = 32x1 + x4 x4 ∈ R
Last update: January 7, 2009 4
