MTH5112 Linear Algebra I 2016–2017
Coursework 1 — Solutions
Exercise 1. In row echelon form: a, c, e, f; in reduced row echelon form: c, f.
Exercise# 2.
(a) The corresponding linear system is
x1 + 3x2
+ 2x4 = 4
x3 + x4 = 3
Leading variables: x1 , x3 ; free variables: x2 , x4 . Now set x2 = α and x4 = β. Using back
substitution we find successively, x3 = 3−x4 = 3−β, and x1 = 4−3x2 −2x4 = 4−3α−2β.
Hence the solution set is {(4 − 3α − 2β, α, 3 − β, β)|α, β ∈ R}.
(b) The corresponding linear system is
x1
x2
x3
= 2
= −1
= 0
Leading variables: x1 , x2 , x3 ; no free variables (i.e. the solution is unique). We can directly
read off the solution and find that the solution set is {(2, −1, 0)}.
(c) The corresponding linear system is
x2
= 7
= −2 .
x3
Leading variables: x2 , x3 ; free variable: x1 . Now set x1 = α, giving the solution set
{(α, 7, −2)|α ∈ R}.
(d) The corresponding linear system is
x1
= 0
0 = 1
0 = 0
Leading variable: x1 ; free variable: x2 . The middle line shows that the system is inconsistent.
Thus the solution set is empty.
Exercise* 3.
(a) A possible example is

1
0

0
0
1
1
0
0

0
0
.
0
0
The matrix has 4 rows and 3 columns and is in row echelon form, since the leading 1 in
the second row is to the right of the leading 1 in the first row. However, the matrix is not
in reduced row echelon form since the leading 1 in the second row is not the only non-zero
element in the second column.
(b) (i) The augmented matrix of the linear system is


1 −1 3 2
 2 −1 0 5
−1 2 −6 0
(ii) Using elementary row operations we find




1 −1 3 2
1 −1 3 2
 2 −1 0 5 ∼ R2 − 2R1 0 1 −6 1
−1 2 −6 0
R3 + R1
0 1 −3 2




1 −1 3 2
1 −1 3 2
0 1 −6 1 ∼
0 1 −6 1 
∼
1
R3 − R2 0 0
3 1
R
0 0
1 13
3 3
1
where the last matrix is now in row echelon form.
(iii) By (b), the linear system corresponding to the row echelon form of the original linear
system is
x1 − x2 + 3x3 = 2
x2 − 6x3 = 1 .
x3 = 13
The leading variables are x1 , x2 , and x3 ; there are no free variables. Using back
substitution starting from the last equation we obtain successively x3 = 31 ; then x2 =
1 + 6x3 = 1 + 6 · 31 = 3; and finally x1 = 2 + x2 − 3x3 = 2 + 3 − 3 · 13 = 4. Thus the
solution set is {(4, 3, 31 )}, i.e. the solution is unique.
Exercise 4. Since p(−1) = a − b + c and p(2) = 4a + 2b + c, we obtain the following 2 × 3 system
for the coefficients a, b, and c:
a − b + c = 1
4a + 2b + c = 10
Applying elementary row operations to the augmented matrix of this system gives:
1 −1 1 1
1 −1 1 1
1 −1 1 1
∼
∼ 1
4 2 1 10
0 6 −3 6
R2 − 4R1
R
0 1 − 12 1
6 2
where the last matrix is now in row echelon form. The corresponding system reads
a − b +
b −
c
1
c
2
= 1
= 1
Here, the leading variables are a and b, and the free variable is c. Now set c = α. Then b = 1+ 12 c =
1 + 12 α and a = 1 + b − c = 1 + (1 + 21 α) − α = 2 − 12 α. Thus the polynomials are of the form
p(x) = (2 − 21 α)x2 + (1 + 12 α)x + α = (2x2 + x) + α2 (−x2 + x + 2), where α is an arbitrary real
number.