Only
linearly independent column
one
A is
3x3
a
Nullity
=>
matrix with
3-1
=
=
a
I dimension
>
-
of
rank
2
1 X 12
-
=
=>
1x
2
1
2-x
-
x3 4x2
+
X
=
=
-
X)(2-x) + 2 +2
4
=
=0
(OB XBI) *
=
-
[A- 1xB-3)E] X
let
XB Xa + 3
From
section (c)
XB
-
0
1A-XA1)1
=>
(1 x) (1
=0
=
=
3
XA
=
(multiple root)
O or 0 or
or
7
4
-
2(1 X)
-
-
(2 x)
-
-
2(1 x)
-
=
0
let A
1
=
(4 , , (3)
(V 31)
-
Set
=>
=
z=C
()
Nullity
=>
=
1
rank
,
=
2
C EIR
,
[
z)"
=> In = (x , X ,
(1
=
2
,
Set
,
4
T
=
=
(
-
EC Ec c1T
,
,
-(34-12) > Solution exists
=
Z =0
E
=>
The
(p
=
(X , y
general
z) (3
-
=
,
solution U
,
2 01T
,
z1T
=
In + Up
=
(3-0 ,+Ec CIT Cer
=
(x , Y ,
,
,
Ux1
·
(1
,
Therefore
2
,
3)
(a , b
,
+ 0
2
=
21 " is not in the vector space
spanned by 1 and E
Null space
For
In
=
C =2
we
,
solution
,
( &C , * C , c)
=
have
basis
a
=
(V , )
:
1-1
=> x
=
a +
=
X 2ax + a b
-
b
(bb)(4) 10)
=
V
=>
=
(1 , 1) , x
(3b)(4) (8)
=
=>
(2
=
(1 ,
-
-
=
x
=
a+
=
1) , X2
=
x
=
y
b
=
y
G-b
=
,
3
(i +)
=
&
-
T
-
Column space of A
det (A x])
does not exist.
0
,
23
det (B
-
xI)
=> X = C
(C X)
=
a
or
-
·
det1A
-
XE )
E
(b
-
V
=
10
0
,
for CA
1), Xi =
,
(bb)(Y) = Y
=
X (1
=
1
,
0
lb
a 2
=>
=
05
,
Xz
,
fr ta
at
=
(b)
,
0) xz
From the results
above
=>
V
1
=
1
=
=
(1
(1 ,
,
-
1 , 0
( 1 ,1
,
0
1
,
,
0
0
=
,
01T M
,
01T N
b
alb
=
=
a+
b
=
a
b
,
,
-
X
,
,
a
for a
-
or
cId
or
e
1
=
1
=
1
=
(0
,
(0
,
(0
,
0
0
0
with all
1
,
,
1
,
0
,
,
0)T Ng
1 ,
0)T N
1
-
,
,
0
in
c+
d
=
c
d
=
e
=
,
,
11T Xs
,
,
-
different from each other
(a)
(a)(2) (2)
Set A
=
E (1 F
=
=
=
A
=
Set X
=> V
=
()
=
(2)
=
( ! (i)
()
sin(A)
=
=
V
sin()
A XXX" XV" Is
=
,
=
UsinCAIN"
=
=
(1)
=
(23)
=-
=
Usin(A)V-
(ii)