CEGEP CHAMPLAIN - ST. LAWRENCE
201-105-RE: Linear Algebra
Patrice Camiré
Problem Sheet #2
Gaussian & Gauss-Jordan Elimination
1. In each case, find two different row-echelon forms of the given matrix.
1 2
−1 1
(a)
(b)
3 4
2 4
2. Would the previous problem be possible if the term ”row-echelon” was replaced by the term ”reduced
row-echelon” ?
3. In each case,
neither.

1 3
(a)  0 1
0 0
determine whether the matrix is in row-echelon form, reduced row-echelon form, or

0
2 
0


1 2 3
(b)  0 0 0 
0 0 0

1 0 0
(c)  0 0 0 
0 0 1
(d)
1 2 0
4
0 0 1 −3

0
 0
(g) 
 0
0

2
(h)  0
0

1
(i)  0
0


1 2 0
(e)  0 1 0 
0 1 0

(f)
0 0 0
0 0 0
1
0
0
0
2
0
0
0
1
1
0
0

0
0
1
0

1
2 

3 
1
0 0
0 0 
0 0

1 1
0 1 
0 0
4. Find the solution set of the linear system of equations. (Use the method of your choice.)
(a)
x + 2y = −3
3x + 4y =
1
(b)
2x + 4y − 6z = −1
5x + 3y + z =
0
3x − y + 7z =
1
x−z = 4
(c) 3x + y − z = 15
x+y+z = 6
(d)
(e)
2x − z = −1
3x + y − z =
1
x + 2y + 3z =
1
2x1 − x3 + 2x4 = −2
3x1 + x2 − x4 =
1
5. Is it possible that a homogeneous linear system of equations has no solutions?
6. Determine whether the homogeneous linear system has one or infinitely many solutions.
(Do as little work as possible.)
πx − y + 3z = 0
(a) √
2x + 2y + 4z = 0
(b)
x1 + 3x2 − 5x3 + 6x4 = 0
2x1 − x2 + 2x4 = 0
5x1 + 4x2 + 3x3 + x4 = 0
(c)
x − 2y + 3z = 0
2x − 3y + 6z = 0
−x + 10y − 2z = 0
(d)
3x + 2y + 5z = 0
x − y + 4z = 0
4x + y + 9z = 0
7. In each case, suppose that the augmented matrix associated to a linear system has been reduced
using elementary row operations to the given row-echelon form. Solve the system using backward
substitution. (Always classify the free variables as parameters, and then solve for the leading
variables.)




1 4
3
1 2
3
9
1 3 −1
2 −6
(a)
0 1 −2
1 −4
7 
(d)  0 1 −4 −5 
(f)  0 0
0 0
1 −1
0 0
0
1
3
1 −5 8


(b)
1
−2
−1
4
−1
0
0 0


 0
1 5 −8 3
1
5
6
2 


(g)
 0
1 2 
1 2
7
0
1 −3
1 
(e)  0 0
(c)
0 0
0 0
0 0 −3
0
0
0
1
1
8. In each case, suppose that the augmented matrix associated to a linear system has been reduced
using elementary row operations to the given reduced row-echelon form. Solve the system. (You
should be able to write down the solution set directly without having to perform any additional
calculations.)




1 0 0 −5
1 0
6 4 0 0 −1
 0 1 −3 7 0 0
8 
8 
(a)  0 1 0

(d) 

0 0 1
3
0 0
0 0 1 0
2 


0 0
0 0 0 1 −10
1 −2 6 7
0 0 0 
(b)  0
1 3 −2 0
1 1
(e)
0
0 0 0
0 0
0 1 −8 2


1 4 0 −1/2
5
1/3 −8 
(c)  0 0 1
0 0 0
0
0
9. Use Gaussian elimination or Gauss-Jordan elimination to solve the following linear systems. (When
row reducing, try to avoid introducing fractions as much as possible.)
(a)
−x + 2y = −2
5x − 11y = 1
(b)
4x + y = −2
3x + 2y = 5
−2x + y − z = 6
(c) −3x − y − z = 1
x − 2y + z = 2
x + 2y + 3z = −4
2x + y + z = 6
(d)
5x + 4y + 5z = 11
(e)
−x + y + 3z = 4
6x − 5y + z = 3
−7x + 6y + 2z = 1
2x1 + x2 − x3 = 3
(f) −x1 − x2 + 3x3 + 4x4 = −7
6x1 + 2x2 + x3 + x4 = −1
4x1 + 7x2 + 4x3
−x1 − x3 + 2x4
(g)
2x1 + 2x2 + 3x3 − 4x4
−x2 − x3 + x4
=
=
=
=
4
−3
2
1
3x1 + x2 − x3 + x4
2x1 + 3x2 + x3 + x4
(h)
x1 − 11x3 + 2x4
−x1 + 6x2 + 2x3 + x4
=
=
=
=
1
−1
2
1
10. For which values of k does the linear system have no solutions, a unique solution, or infinitely many
solutions?
(a)
kx − 2y = 3k + 2
x + (k − 3)y = 4
−2x + (k − 3)y + kz = k + 11
(k + 1)y − z = 5
(b)
x + 2y + z = −1
x + 2y − 3z = 4
3x − y + 5z = 2
(c)
4x + y + (k 2 − 14)z = k + 2
(k 2 − 9)x = k + 3
(d) (k − 5)y = 1
(k + 8)z = −10
11. (a) Find the equation of the cubic polynomial passing through the points (0, −1), (1, 0), (−1, −4)
and (2, 5).
(b) Show that the equation of a circle can be expressed in the form x2 + y 2 + ax + by = c, for some
real numbers a, b, c ∈ R.
(c) Explain why three points on a circle are required to determine its equation. Find the equation
of the circle passing through the points (0, −1), (2, 1) and (3, −1). Provide its center and
radius.
12. Use a proper change of variables to solve the given nonlinear system of equations. How many
solutions are there?
1
z
1
2
x
(b) −2e + 2 +
y
z
1
3
4ex − 2 −
y
z
ex −
(a)
−2 sin(θ) + 5x2 = 45
, where 0 ≤ θ ≤ 2π.
− sin(θ) + 2x2 = 18
= 8
= −12
= 23
13. (Optional)
(a) Show that if ad − bc 6= 0, then the reduced row-echelon form of
(b) Show that if ad − bc 6= 0, then the linear system
any e, f ∈ R.
a b
c d
is
1 0
.
0 1
ax + by = e
has a unique solution for
cx + dy = f
Answers
1. (a)
1 2
0 1
and
1 4/3
0
1
(b)
1 −1
0
1
and
1 2
0 1
2. No. The reduced row-echelon form of a matrix is unique.
3. (a) r.e.f.
(b) r.r.e.f.
(c) neither
x = 7
4. (a)
y = −5
(b)



x =




3
− 11t
14
−5
+ 16t ; t ∈ R
14
y =






z = 7t
(d) r.r.e.f.
(g) r.e.f.
(e) neither
(h) neither
(f) r.r.e.f.
(i) r.e.f.
(c) ∅

x1



x2
(d)
x


 3
x4

 x
y
(e)

z
=
=
=
=
−1 + r − s
4 − 3r + 4s
; r, s ∈ R
2r
s
= −6/5
= 16/5
= −7/5
5. No. A homogeneous linear system always has at least the trivial solution x1 = x2 = · · · = xn = 0.
6. (a) The system has infinitely many solutions. (No work is required.)
(b) The system has infinitely many solutions. (No work is required.)
(c) The system has only the trivial solution. (Work is required.)
(d) The system has infinitely many solutions. (Work is required.)

x = 11
x1 = 7 − 3t


7. (a)

y = −2
x2 = t
(f)
; t∈R
x3 = 19

x = 8 + 5t


(b)
; t∈R
x4 = 3
y = t

(c) ∅
x1 = −49




x2 = −24
 x = 30
(g)
x = 4

y = −9
(d)

 3

x4 = 1
z = −1

 x = 19 − 5t
y = t
(e)
; t∈R

z = 2
8. (a)
(b)
(c)
9. (a)
(b)
(c)

 x
y

z

 x
y

z


 x1

x2
x


 3
x4
x
y
x
y

 x
y

z
= −5
= 8
= 3
(d)
= 7 + 2s − 6t
= s
; s, t ∈ R
= t
=
=
=
=
5 + 3t − 4s
s
; s, t ∈ R
−8 − 2t
6t
(e)
= 20
= 9
(f)
= −9/5
= 26/5
= −21
= 13
= 49
(d) ∅

 x =
y =
(e)

z =

 ∅ :
1 :
10. (a)

∞ :

 ∅ :
1 :
(b)

∞ :
(g)
23 − 16t
27 − 19t ; t ∈ R
t
(h)
k=1
k ∈ R − {1, 2}
k=2
(c)
k = −3, −1
k ∈ R − {−3, −1}
never
(d)

x1




x2



x3
x4





x

 5
x6

x1




 x2
x3


x


 4
x5

x1



x2
x


 3
x4

x1



x2
x


 3
x4

x1



x2
x


 3
x4

 ∅
1

∞

 ∅
1

∞
=
=
=
=
=
=
−1 − 6s − 4t
8 + 3s − 7t
s
; s, t ∈ R
t
2
−10
=
=
=
=
=
1 − 3r + 2s − t
r
s
; r, s, t ∈ R
2 + 8t
t
=
=
=
=
−2 + 10t
6 − 27t
; t∈R
−1 − 7t
t
=
=
=
=
1
−16
28
13
=
=
=
=
26
34
−22
−133
: k = −4
: k ∈ R − {−4, 4}
: k=4
: k = −8, 3, 5
: k ∈ R − {−8, −3, 3, 5}
: k = −3
11. (a) y = x3 − x2 + x − 1
(b) The equation of a circle with center (xo , yo ) and radius r is naturally represented by the
equation (x − xo )2 + (y − yo )2 = r2 . If we expand both squares and send the constant terms to
the right side of the equality, the equation becomes x2 +y 2 +(−2xo )x+(−2yo )y = r2 − x2o − yo2 .
| {z }
| {z }
|
{z
}
(c) x2 + y 2 − 3x + y = 0, with center
12. (a)
3 1
,−
2 2
√
10
and radius
.
2
θ = 0, π, 2π
, and so there are six solutions.
x = −3, 3

 x = ln(3)
y = −1/2, 1/2 , and so there are two solutions.
(b)

z = −1/5
13.
=a
=b
=c