6CCM113A Linear Methods, Autumn 20010
Solutions to QS 4
1. Notice that the first two planes are the same in all three systems.
Solution of (1). We put the augmented matrix into reduced row echelon form:






1 1 1 2
1 1 1 2
1 1 1
2
R3 +R1
−2R2
3
 0 2 1 0  R2 ↔(1/2)R
 0 1 0 −1  R3−→
 0 2 1
−→
0  −→
0 2 0 −2
0 2 1 0
−1 1 −1 −4

1 1 1
0 1 0
0 0 1


1
2
R1 −R3
0
−1  −→
0
2
1
1
0
0
0
1


1 0
0
R1 −R2
0 1
−1  −→
0 0
2
0
0
1

1
−1 
2


1
⇒ x =  −1 
2
These three planes intersect in a single point (unique solution).
The intersection of the first two planes is a straight line (Why? Hint: Look at the first two
rows of the REF) which intersects the third plane in a single point.
Solution of (2).
reduction we obtain,

1 1 1
 0 2 1
−1 1 0
Beginning with the augmented matrix, after two steps of row


1
2


→ ··· → 0
0
0
−4
1
2
0
1
1
0

2
0 
−2
⇒ inconsistent
These planes have no common points of intersection (no solution).
The intersection of the first two planes is a straight line which is parallel to the third plane
and does not intersect it.
Solution of (3). This time the row reduction leads to a matrix with a row of zeros,




1 1 1
2
1 0 21
2




0  → · · · →  0 1 12
⇒ Set z = t
0
 0 2 1
−1 1 0
−2
0 0 0
0
Then,
 
 1
2
−2

 
 1
1 
x =  0 − 2t  =  0  + t  −2 
0+t
0
1

2 − 12 t

These planes intersect in a line (infinitely many solutions).
The intersection of the first two planes is a straight line which is contained in the third
plane.
2. NO: The two equations represent a pair of planes in R3 which means their intersection will
either be a line (this happens when the planes have non-parallel normal vectors) or the
1
two planes are parallel planes with empty intersection or two coincident planes with two
dimensional solution set.
3.


0
0
1

1

−→
0
0
1
Similarly the others all reduce further except for
0
1
0
0
R3 *
)R2

0
1
0
2
which is in REF.
0
4. Put the augmented matrix into REF form:

1
 −1
1
−1
1
−1

1
R3 −R2

−→
0
0
1 1 2
0 1 −1
2 3 4
−1
0
0
1 1 2
1 2 1
0 0 1

1

−→
0
0
R1 −R2


1
4
R2 +R1

0
−3  R−→
3 −R1
−→
0
7

4
1
2
−1
0
0

R1 −2R3
−→
−→
1
0
0
0 −1
1 2
0 0
0
0
1
R2 −R3
−1 1 1 2
0 1 2 1
0 1 2 2

4
1
3
−1
0
0

0
−1 
2
1 1 0
1 2 0
0 0 1

1
−1 
2
Set the non-leading variables to arbitrary constants: x2 = s, x4 = t, and solve for the
leading variables in terms of these parameters, starting with the bottom row,
=⇒ x5 = 2,
x4 = t,
x3 = −1 − 2t,
x2 = s,
x1 = 1 + s + t



 
 
 
1
1
1
x1
1+s+t
s
 0 
1
 x2  
  0 


 

  
 
=⇒ x =  x3  =  −1 − 2t  =  −1  + s  0  + t  −2  = p + su + tv



 
  
 
0
1
t
0
x4
0
0
2
2
x5

All you need to do now is to verify that p, u, v satisfy the desired conditions. Which is
straightforward. It will be more instructive to solve both systems side-by-side.
Finally the answer to the last question is no. No matter what the last column it it impossible
for the REF of the augmented matrix to contain a row like (0, 0, 0, 0, 0, 1). The reason is
that irrespective of what the last column is all the leading 1s are already occurring outside
the last column.
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