Stony Brook University
Introduction to Linear Algebra
Mathematics Department
MAT 211
Julia Viro
March 10, 2009
Solutions for Midterm I
Problem 1.
Solve the system using the GaussJordan elimination and verify your answer
x
1
+ 2
x
3

x
4
=1
x
2
+ 2
x
4
=

1
x
1

x
2
+ 2
x
3

3
x
4
=2
.
Solution.
Elementary row transformations of the augmented matrix of the system give rise to
the reduced rowechelon form:
1
0 2

1
1
0
1 0
2

1
1

1 2

3
2
∼
R
3

R
1
1
0 2

1
1
0
1 0
2

1
0

1 0

2
1
∼
R
3+
R
2
1 0 2

1
1
0 1 0
2

1
0 0 0
0
0
.
Obviously, the rank of the matrix is 2. The number of free variables is 4

2 = 2 (the number
of unknowns minus the rank). We write down the solution starting from the back:
x
4
=
t
(choose
x
4
as a free variable),
x
3
=
s
(choose
x
3
as a free variable),
x
2
=

1

2
t
(from the second row in the rref),
x
1
= 1 +
t

2
s
(from the ﬁrst row in the rref).
So the solution is
(
x
1
,x
2
,x
3
,x
4
) = (1 +
t

2
s,

1

2
t,s,t
) = (1
,

1
,
0
,
0) +
t
(1
,

2
,
0
,
1) +
s
(

2
,
0
,
1
,
0)
,
where
t
and
s
are arbitrary real numbers. Geometrically, the solution is a plane in
R
4
.
To verify the solution, we substitute it into the three equations of the system:
1 +
t

2
s
+ 2
s

t
= 1

1

2
t
+ 2
t
=

1
1 +
t

2
s
+ 1 + 2
t
+ 2
s

3
t
= 2
.
Since all the equations are satisﬁed, our solution is correct.
Answer:
(
x
1
,x
2
,x
3
,x
4
) = (1 +
t

2
s,

1

2
t,s,t
)
, t, s
∈
R
.