Let’s solve it. Subtracting the first equation from the second one we get
2
a
= 3
m

1
,
so
a
=
3
m

1
2
.
Now,
b
= 1

a
= 1

3
m

1
2
=
3

3
m
2
.
Now, the matrix form of the approximation equation is
aA
+
bI
=
A
m
.
Substituting
a, b
and
A
, we get:
A
m
=
3
m

1
2
ˆ
2
1
1
2
!
+
3

3
m
2
ˆ
1
0
0
1
!
=
1
2
ˆ
2(3
m

1) + (3

3
m
)
3
m

1
3
m

1
2(3
m

1) + (3

3
m
)
!
=
1
2
ˆ
3
m
+ 1
3
m

1
3
m

1
3
m
+ 1
!
.
3. Find the distance between the vector
u
= (10
,
5
,

12) and the subspace, spanned by the following
vectors:
v
1
= (4
,
1
,
2);
v
2
= (5
,
8
,
7);
v
3
= (6
,

3
,
0)
.
Solution:
First we should apply the GramSchmidt orthogonalization process to vectors
v
1
, v
2
,
and
v
3
:
w
1
=
v
1
= (4
,
1
,
2);
w
2
=
v
2

h
v
3
, w
1
i
h
w
1
, w
1
i
w
1
= (5
,
8
,
7)

20 + 8 + 14
16 + 1 + 4
(4
,
1
,
2)
= (5
,
8
,
7)

2(4
,
1
,
2)
= (

3
,
6
,
3);
w
3
=
v
3

h
v
3
, w
1
i
h
w
1
, w
1
i
w
1

h
v
3
, w
2
i
h
w
2
, w
2
i
w
2
= (6
,

3
,
0)

24

3
16 + 1 + 4
(4
,
1
,
2)


18

18
9 + 36 + 9
(

3
,
6
,
3)
= (6
,

3
,
0)

(4
,
1
,
2) +
2
3
(

3
,
6
,
3)
= (0
,
0
,
0)
.
3