TEXT 2.1
# 12
We set x1 = t and solve for x2.
Setting x1 = t yields 2t + 3x2 = 5. Solving for x2 yields 3x2 = 5 – 2t  x2 =
So, a complete set of solution written in parametric form is [t,
5 2
 t.
3 3
5 2
 t ].
3 3
Note: We could have set x2 = t and solved for x1 to get the parametric [
5 3
 t , t].
2 2
# 16
The lines intersect at (3, –2), so the unique solution is [3, –2]. To solve, subtract 3×1st from 2nd
 7y = –14  y = –2, so substitution  x = 3.
# 22
We find the solution [x1, x2, x3] = [0, 0, 0] using back substitution.
Note: This follows immediately from the fact that all three equations are equal to zero.
TEXT 2.1
#8
This matrix is in row echelon form, but not reduced row echelon form. Why not?
The leading entry in row 4 is not a 1. Could we have given another reason?
# 18
R1+R2
 2 0  1


A =  1 1 0
 1 1
1


R2+2R3
 3 1  1


 1 1 0
 1 1
1


R3+R1
 3 1  1


2
 1 3
 1 1
1


 3 1  1


2
 1 3
 1 2
0

R2+ 12 R3
R2+R1
 3 1  1
 3 1  1


 2 4 1   3 5 1 = B
2 2
2 2
0
0


Therefore, the matrices A and B are row equivalent.
# 26
We form the augmented matrix and row reduce it as follows:
R3 – 5R1 –2 R2
0
 1  1 1 0
 1 1 1



  1 3 1 5
3 1 5
 1
 0
 3
0 0  8
1 7 2


The third row is equivalent to the equation 0 = –8 which clearly has no solution.
Therefore, the system in inconsistent.
Does R3 = 5R1 + 2 R2 (excluding constants) cause the system to be inconsistent.
# 34
When there are 4 equations and 4 variables, if the solution exists it is unique.
Why? Because the Rank Theorem (2.2) tells us there are 4 – 4 = 0 free variables.
1

1
1

1
4
1 0


2 3 4 10
0 1
→ 
0 0
3 6 10 20



0 0
4 10 20 35
1
1
1
0 0 1

0 0 1
 The solution is
1 0 1

0 1 1
 a  1
 b  1
  =  .
 c  1
  
d  1
# 40
First row reduce the system [A│x] and then answers parts (a), (b) and (c).
R1 ↔ R2
R2 – 12 kR1
2
3
k


 2  4  6

 2  4  6


2
3
k

4
 6
2


0 2  2k 3  3k 
(a) There are no values of k for which this system has no solution. Why?
The system has no solution when A has a zero row with corresponding constant ≠ 0.
2+ 2k = 0  k = –1 is the only value of k that creates a row of all zeros.
But k = –1  the constant 3 + 3k = 3 – 3 = 0. What does this imply?
(b) The system has a unique solution for k ≠ –1. Why?
From (a), we see when k ≠ –1 then rank A = 2. So, there are 2 – 2 =0 free variables.
(c) The only value of k for which system infinitely many solutions is k = –1.
The system has infinitely many solutions when A has a zero row with constant = 0.
From (a), we see this is exactly the case when k = –1.
# 42
First row reduce the system [A│x] and then answers parts (a), (b) and (c).
R3 – R1 – R2
 1  2 3 2
 1  2 3 2




1 1 k 
1 1 k
1
1




2  1 4 k 2 
2  1 4 k 2 
(a) When k ≠ –1, 2, this system has no solution. Why?
The system has no solution when A has a zero row with corresponding constant ≠ 0.
k2 – k – 2 = 0  k = –1, 2 are the only values of k that that make the constant 0.
(b) This system never has a unique solution because rank A = 2. Why?
Since rank A = 2, there is always at least 3 – 2 = 1 free variable.
(c) When k = –1, 2, this system infinitely many solutions.
The system has infinitely many solutions when A has a zero row with constant = 0.
From (a), we see this is exactly the case when k = –1, 2.
# 42
(a) The following system of n equation has infinitely many solutions:
x1 + x2 + … + xn = 0
2x1 + 2x2 + … + 2xn = 0

nx1 + nx2 + … + nxn = 0
Likewise, the following system of n +1 equation has infinitely many solutions:
x1 + x2 + … + xn = 0
2x1 + 2x2 + … + 2xn = 0

nx1 + nx2 + … + nxn = 0
(n+1) x1 + (n+2) x2 + … + (n+1) xn = 0
(b) The system of n equation x1 = 0, x2 = 0, … , xn = 0 has the unique solution xi = 0.
As does the system of 2n equations x1 = 0, x2 = 0, … , xn = 0.