Determining Eigenvalues and Eigenvectors for a Given Matrix

The assignment consists of six questions related to linear algebra, specifically involving matrices and eigenvalues. Questions cover topics such as finding the eigenvalues and eigenvectors of a given matrix, checking if a computed matrix is correct based on its eigenvalues, and determining the condition of a matrix based on its properties.

SIT292 LINEAR ALGEBRA 2017
ASSIGNMENT 2
STUDENT ID
[Pick the date]

Question 1
(i) Given matrix
A=
[ 3 2 1
1 1 2
−1 −2 0 ]
 Cofactors Ci jof the matrix
Ci j=¿
C11=¿
C12 =¿
C13 =¿
C21 =¿
C22 =¿
C23 =¿
C31 =¿
C32 =¿
C33 =¿
Hence, cofactors of the matrix A would be given below:
1

Cofactors A= [ 4 −2 −1
−2 1 4
3 −5 1 ]
 Adj Aof the matrix A would be given below:
Adj A= [ 4 −2 3
−2 1 −5
−1 4 1 ]
(ii) Verification that computed adj Ais correct.
If A∗( adj A )
det A =I
Now,
LHS
A∗( adj A )= [ 3 2 1
1 1 2
−1 −2 0 ]. [ 4 −2 3
−2 1 −5
−1 4 1 ]
¿ [ 3.4+2. (−2)+1.(−1) 3.(−2)+2.1+1.4 3.3+ 2.(−5)+ 1.1
1.4+1.(−2)+2.(−1) 1.(−2)+1.1+2.4 1.3+ 1.(−5)+ 2.1
−1.4 +(−2) .(−2)+0.(−1) ( −1 ) . ( −2 ) +(−2).1+ 0.4 (−1).3+(−2). (−5)+ 0.1 ]
¿ [ 7 0 0
0 7 0
0 0 7 ]
And
A=
[ 3 2 1
1 1 2
−1 −2 0 ] 2

det A=
| 3 2 1
1 1 2
−1 −2 0|
¿ 3 ( 0+ 4 ) −2 ( 0+ 2 )+ 1 (−2+1 )
¿ 12−4−1
¿ 7
A∗( adj A )
det A =
[ 7 0 0
0 7 0
0 0 7 ]7 = [ 1 0 0
0 1 0
0 0 1 ]
RHS
I =
[1 0 0
0 1 0
0 0 1 ]
LHS = RHS
It is apparent that both the sides are equal and therefore, the computed adj A is correct.
3

Question 2
Two vectors would be orthogonal when there dot product becomes zero.
¿ 7 ∝+ ∝. ∝+3∝ . (−3 ) +1.1+ (−4 ) .4
¿ 7 ∝+ ∝2 −9 ∝+1−16
¿ ∝2−2 ∝−15
Hence,
∝2−2 ∝−15=0
( ∝−5 ) ( ∝+3 )=0
4

∝=5 ,−3
Therefore, the for ∝=5 ,−3the vectors would be orthogonal .
Question 3
Given equations
Gaussian elimination method to reduce the following system of equations into row echelon form
is applied below:
5

The row echelon form of the given system is given below:
This system does not have any solution because 0 ≠−5 .
It is apparent from the above that the given system has three linear equations and 4 variables.
Hence, the system is said to be inconsistent and does not have any solution as evident from the
above.
6

Question 4
Eigen values and eigenvectors of the given matrixes are as highlighted below:
 For matrix A
Eigen values
A=
[1 0 1
0 1 0
1 0 1 ]
Let λ is the eigenvalues of the given matrix in such a way that
det ( A−¿ λI )=0 ¿
Now,
A−λI = [1 0 1
0 1 0
1 0 1 ]−λ [1 0 0
0 1 0
0 0 1 ]
A−λI = [ 1 0 1
0 1 0
1 0 1 ] −
[|
λ 0 0
0 λ 0
0 0 λ|]
A−λI = [ 1−λ 0 1
0 1−λ 0
1 0 1−λ ]
7

Determinant of A−λI
|A−λI |=
|1−λ 0 1
0 1−λ 0
1 0 1− λ|
¿ ( 1− λ ) { 0− ( 1−λ ) ( 1−λ ) }−0 { 0−0 }+1 { 0− ( 1−λ ) }
¿ ( 1− λ ) ¿
¿ ( 1− λ ) ¿
¿ ( 1− λ ) { ( λ−1 )2−1 }
¿ ( 1− λ ) { λ2−2 λ+ 1−1 }
¿ ( 1− λ ) ( λ2−2 λ )
det ( A−¿ λI )= λ ( 1−λ ) ( λ−2 ) ¿
Now,
det ( A−¿ λI )=0 ¿
λ ( 1− λ ) ( λ−2 ) =0
λ=0 ,
1− λ=0
λ=1 ,
λ−2=0
λ=2
8

Therefore, the eigenvalues of matrix A is 0,1,2.
Eigenvectors
Let the eigenvector is ϑ, such that ( A−λ I ) ϑ =0 for each of the corresponding eigenvalue.
Now,
For eigenvalue λ=0 ,
( A−0 I ) ϑ =0
( [1 0 1
0 1 0
1 0 1 ]−0 [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]= [0
0
0 ]
[1 0 1
0 1 0
0 0 0 ][ x
y
z ]= [0
0
0 ]
x + z=0
y=0
Now, x=−z
y=0
z=z
Eigenvector
[ x
y
z ]= [−z
0
z ] 9

Let z=1∧hence ,
Eigenvector for λ=0 is [−1
0
1 ]
For eigenvalue λ=1 ,
( A−1 I ) ϑ=0
( [1 0 1
0 1 0
1 0 1 ]−1 [1 0 0
0 1 0
0 0 1 ]) [ x
y
z ]= [0
0
0 ]
[0 0 1
0 0 0
1 0 0 ][ x
y
z ]= [0
0
0 ]
Now reduce the matrix into row echelon form as
[a … . b
0 … . …
0 0 c ]
[0 0 1
0 0 0
1 0 0 ]: [1 0 0
0 0 0
0 0 1 ]
Further,
Reduce the matrix to reduced row echelon form
[1 0 b
0 0 0
0 0 1 ] 10

[1 0 0
0 0 0
0 0 1 ]: [1 0 0
0 0 1
0 0 0 ]
Hence,
[1 0 0
0 0 1
0 0 0 ][ x
y
z ]= [0
0
0 ]
x=0
z=0
Eigenvector
[ x
y
z ]= [ 0
y
0 ]= [0
0
0 ]
Let y=1∧hence ,
Eigenvector for λ=1 , is [ 0
1
0 ]
Eigenvector for λ=2
( A−2 I ) ϑ=0
( [1 0 1
0 1 0
1 0 1 ]−2 [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]= [0
0
0 ]
11

[−1 0 1
0 −1 0
1 0 −1 ][ x
y
z ]= [0
0
0 ]
Reduce the above matrix as
[a … b
0 .. …
0 0 c ]
[−1 0 1
0 −1 0
0 0 0 ]
Now reduce the matrix into reduced row echelon form as
[1 … b
0 .. …
0 0 1 ]
[1 0 −1
0 1 0
0 0 0 ]
[1 0 −1
0 1 0
0 0 0 ][ x
y
z ]= [0
0
0 ]
x−z=0
y=0
Now, x=z
y=0
12

z=z
Eigenvector
[ x
y
z ]= [ z
0
z ]
Let z=1∧hence ,
Eigenvector for λ=2 is [ 1
0
1 ]
Therefore, the eigenvectors of matrix A is [−1
0
1 ], [0
1
0 ], [1
0
1 ]
 For matrix B
Eigen values
B= [3 2 1
3 2 1
3 2 1 ]
Let λ is the eigenvalues of the given matrix in such a way that
det (B−¿ λI)=0 ¿
Now,
B−λI = [ 3 2 1
3 2 1
3 2 1 ]− λ [ 1 0 0
0 1 0
0 0 1 ]
13

B−λI = [ 3 2 1
3 2 1
3 2 1 ]−
[|
λ 0 0
0 λ 0
0 0 λ|]
B−λI = [ 3− λ 2 1
3 2−λ 1
3 2 1−λ ]
Determinant of B−λI
|B− λI|=
|3−λ 2 1
3 2−λ 1
3 2 1−λ |
¿ ( 3−λ ) ( λ2 −3 λ ) −2 (−3 λ ) +1.(3 λ)
¿− ( λ−6 ) λ2
det (B−¿ λI)=− ( λ−6 ) λ2 ¿
Now,
det (B−¿ λI)=0 ¿
− ( λ−6 ) λ2=0
( λ−6 ) λ2=0 ,
λ=0 ,
λ−6=0
λ=6
Therefore, the eigenvalues of matrix B is 0 , 6 .
14

Eigenvectors
Let the eigenvector is ϑ, such that ( B−λ I ) ϑ=0 for each of the corresponding eigenvalue.
Now,
For eigenvalue λ=0 ,
( B−0 I ) ϑ=0
( [3 2 1
3 2 1
3 2 1 ]−0 [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]= [0
0
0 ]
[3 2 1
3 2 1
3 2 1 ][ x
y
z ]= [0
0
0 ]
Now reduce the matrix into row echelon form as
[a … . b
0 … . …
0 0 c ]
[3 2 1
3 2 1
3 2 1 ]: [3 2 1
0 0 0
0 0 0 ]
Now reduce the matrix into reduced row echelon form as
[1 0 b
0 0 0
0 0 1 ]
[ 3 2 1
0 0 0
0 0 0 ] :
[ 1 2
3
1
3
0 0 0
0 0 0 ] 15

Hence,
[ 1 2
3
1
3
0 0 0
0 0 0 ] [ x
y
z ]= [ 0
0
0 ]
x +( 2
3 ) y +( 1
3 ) z=0
x=− ( 2
3 ) y−( 1
3 ) z
Now,
Eigenvector
[ −( 2
3 ) y− ( 1
3 ) z
y
z ]=
[− ( 2
3 ) y
y
0 ] +
[− ( 1
3 ) z
0
z ] where , y∧z ≠ 0
Let y∧z =1
[−( 2
3 )
1
0 ] ,
[− ( 1
3 )
0
1 ]
Eigenvector for λ=0 is
[ −( 2
3 )
1
0 ] ,
[− ( 1
3 )
0
1 ]
For eigenvalue λ=6 ,
( B−0 I ) ϑ=0
16

( [3 2 1
3 2 1
3 2 1 ]−6 [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]=
[0
0
0 ]
[ −3 2 1
3 −4 1
3 2 −5 ][ x
y
z ] = [ 0
0
0 ]
Now reduce the matrix into row echelon form as
[a … . b
0 … . …
0 0 c ]
[−3 2 1
3 −4 1
3 2 −5 ]: [−3 2 1
0 4 −4
0 0 0 ]
Now reduce the matrix into reduced row echelon form as
[1 0 b
0 0 0
0 0 1 ]
[ −3 2 1
0 4 −4
0 0 0 ] : [ 1 0 −1
0 1 −1
0 0 0 ]
Hence,
[1 0 −1
0 1 −1
0 0 0 ][ x
y
z ]= [0
0
0 ]
x−z=0
y−z =0
17

Now,
x=z
y=z
[ x
y
z ]= [ z
z
z ]
z ≠ 0∧hence let z=1
[ z
z
z ]= [1
1
1 ]
Eigenvector for λ=6 is [1
1
1 ]
Question 5
(a) Eigen values and eigenvectors of the given matrixes
Given
A=
[1 1 4
0 −1 1
0 0 2 ]
Let λ is the eigenvalues of the given matrix in such a way that
det ( A−¿ λI )=0 ¿
Now,
18

A−λI = [1 1 4
0 −1 1
0 0 2 ]−λ [1 0 0
0 1 0
0 0 1 ]
A−λI = [ 1 1 4
0 −1 1
0 0 2 ] − [ λ 0 0
0 λ 0
0 0 λ ]
A−λI = [1−λ 1 4
0 −1−λ 1
0 0 2−λ ]
Determinant of A−λI
| A−λI |=
|
1−λ 1 4
0 −1−λ 1
0 0 2−λ|
¿ ( 1− λ ) (λ2−λ−2)−1 {0−0 }+ 4 { 0−0 }
¿ ( 1− λ ) (λ−2)(λ+1)
det ( A−¿ λI )= ( 1−λ ) ( λ−2)( λ +1)¿
Now,
det ( A−¿ λI )=0 ¿
( 1− λ ) ( λ−2 ) ( λ+1 ) =0
1− λ=0 ,
λ=1 ,
λ−2=0
19

λ=2 ,
λ+1=0
λ=−1
Therefore, the eigenvalues of matrix A is 1 ,2 ,−1.
Eigenvectors
Let the eigenvector is ϑ, such that ( A−λ I ) ϑ =0 for each of the corresponding eigenvalue.
Now,
For eigenvalue λ=1 ,
( A−1 I ) ϑ=0
( [1 1 4
0 −1 1
0 0 2 ]−1 [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]= [0
0
0 ]
[ 0 1 4
0 −2 1
0 0 1 ][ x
y
z ] =
[ 0
0
0 ]
Now reduce the matrix into row echelon form as
[1 … . b
0 … . …
0 0 1 ]
20

[ 0 1 4
0 −2 1
0 0 1 ] : [ 0 1 0
0 0 0
0 0 1 ]
Hence,
[0 1 0
0 0 0
0 0 1 ][ x
y
z ]= [0
0
0 ]
y=0
z=0
Now,
[ x
y
z ]= [ x
0
0 ]
Let x=1∧hence ,
Eigenvector for λ=0 is [1
0
0 ]
For eigenvalue λ=2 ,
( A−2 I ) ϑ=0
( [1 1 4
0 −1 1
0 0 2 ]−2 [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]=
[0
0
0 ]
21

[−1 1 4
0 −3 1
0 0 0 ][ x
y
z ]=
[0
0
0 ]
Now reduce the matrix into row echelon form as
[1 … . b
0 … . …
0 0 1 ]
[−1 1 4
0 −3 1
0 0 0 ]: [−1 1 4
0 −3 1
0 0 0 ]
Now reduce the matrix into reduced row echelon form as
R 2← (−1
3 ) R 2
[ −1 1 4
0 1 −1
3
0 0 0 ]
R 1← R 1− ( 1. R 2 )
[ −1 0 13
3
0 1 −1
3
0 0 0 ]
R 1←−R1
[1 0 −13
3
0 1 −1
3
0 0 0 ] 22

Hence,
[1 0 −13
3
0 1 −1
3
0 0 0 ].
[ x
y
z ]= [0
0
0 ]
x− (13
3 )z =0
y− ( 1
3 ) z=0
And,
x= (13
3 )z ∧ y=( 1
3 )z
Now,
[ x
y
z ]=
[ ( 13
3 ) z
( 1
3 ) z
z ]
Let z=1∧hence ,
Eigenvector for λ=2 is
[ ( 13
3 )
( 1
3 )
1 ]
23

For eigenvalue λ=−1 ,
( A−(−1)I ) ϑ =0
( [1 1 4
0 −1 1
0 0 2 ]−(−1) [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]=
[0
0
0 ]
[2 1 4
0 0 1
0 0 3 ][ x
y
z ]= [0
0
0 ]
Now reduce the matrix into row echelon form as
[1 … . b
0 … . …
0 0 1 ]
[ 2 1 4
0 0 1
0 0 3 ] :
[ 1 1
2 0
0 0 1
0 0 0 ]
Hence,
[1 1
2 0
0 0 1
0 0 0 ] [ x
y
z ]= [0
0
0 ]
x +( 1
2 ) y=0
z=0
Now,
24

x=− ( 1
2 ) y ∧z=0
[ x
y
z ]=
[−( 1
2 ) y
y
0 ]
Let y=1∧hence ,
Eigenvector for λ=−1 is
[ −( 1
2 )
1
0 ]
 For matrix B
B= [2 2 1
0 1 3
0 0 −1 ]
Let λ is the eigenvalues of the given matrix in such a way that
det (B−¿ λI)=0 ¿
Now,
B−λI = [2 2 1
0 1 3
0 0 −1 ]−λ [1 0 0
0 1 0
0 0 1 ]
B−λI = [2 2 1
0 1 3
0 0 −1 ]− [λ 0 0
0 λ 0
0 0 λ ] 25

B−λI = [2−λ 2 1
0 1− λ 3
0 0 −1−λ ]
Determinant of B−λI
|B− λI|=
|2−λ 2 1
0 1−λ 3
0 0 −1−λ |
¿ ( 2− λ ) (λ2−1)−2 { 0 } +1 { 0−0 }
¿ ( 2− λ ) (λ2−1)
det ( B−¿ λI )= ( 2−λ ) (λ2−1)¿
Now,
det (B−¿ λI)=0 ¿
( 2− λ ) (λ2−1)=0
( 2− λ ) ( λ−1 ) ( λ+1 ) =0 ,
λ=1 ,−1 , 2
Therefore, the eigenvalues of matrix A is 1 ,−1 , 2.
Eigenvectors
Let the eigenvector is ϑ, such that ( A−λ I ) ϑ =0 for each of the corresponding eigenvalue.
Now,
For eigenvalue λ=1 ,
26

( B−1 I ) ϑ =0
( [ 2 2 1
0 1 3
0 0 −1 ] −1 [ 1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ] =
[ 0
0
0 ]
[ 1 2 1
0 0 3
0 0 −2 ][ x
y
z ] = [ 0
0
0 ]
Now reduce the matrix into row echelon form as
[1 … . b
0 … . …
0 0 1 ]
[ 1 2 1
0 0 3
0 0 −2 ] : [ 1 2 0
0 0 1
0 0 0 ]
Hence,
[1 2 0
0 0 1
0 0 0 ][ x
y
z ]= [0
0
0 ]
x +2 y =0 and x=−2 y
z=0
Now,
[ x
y
z ]= [−2 y
y
z ]
Let y=1∧hence ,
27

Eigenvector for λ=0 is [−2
1
0 ]
For eigenvalue λ=−1
( B−(−1) I ) ϑ=0
( [2 2 1
0 1 3
0 0 −1 ]−(−1) [1 0 0
0 1 0
0 0 1 ]) [ x
y
z ]= [0
0
0 ]
[3 2 1
0 2 3
0 0 0 ][ x
y
z ]= [0
0
0 ]
Now reduce the matrix into row echelon form as
[1 … . b
0 … . …
0 0 1 ]
[3 2 1
0 2 3
0 0 0 ]:
[1 0 −2
3
0 1 3
2
0 0 0 ]
Hence,
[ 1 0 −2
3
0 1 3
2
0 0 0 ] [ x
y
z ]= [ 0
0
0 ]
28

x− 2
3 z=0∧x= 2
3 z
y +( 3
2 )z=0∧ y =(−3
2 ) z
Now,
[ x
y
z ]=
[ 2
3 z
( −3
2 ) z
z ]
Let z=1∧hence ,
Eigenvector for λ=−1 is
[ 2
3
( 3
2 )
1 ]
For eigenvalue λ=2
( B−(2)I ) ϑ=0
( [ 2 2 1
0 1 3
0 0 −1 ] −(2) [ 1 0 0
0 1 0
0 0 1 ]) [ x
y
z ] = [ 0
0
0 ]
[ 0 2 1
0 −1 3
0 0 −3 ][ x
y
z ] = [ 0
0
0 ]
Now reduce the matrix into row echelon form as
29

[1 … . b
0 … . …
0 0 1 ]
[ 0 2 1
0 −1 3
0 0 −3 ] : [ 0 1 0
0 0 0
0 0 1 ]
Hence,
[0 1 0
0 0 0
0 0 1 ][ x
y
z ]= [0
0
0 ]
y=0
z=0
Now,
[ x
y
z ]= [ x
0
0 ]
Let x=1∧hence ,
Eigenvector for λ=2 is [ 1
0
0 ]
(b) Condition P−1 AP=B
Given matrix
A = [1 1 4
0 −1 1
0 0 2 ] 30

B= [2 2 1
0 1 3
0 0 −1 ]
In this case, the above condition would be fulfilled only when P would be and invertible matrix
and matrix B would be a diagonal matrix.
However, it is apparent from the above that matrix B is not a diagonal matrix and hence, matrix
P would not be determined for the condition P−1 AP=B .
Question 6
Given matrix
31

A = [3 2 −1
3 8 −3
3 6 −1 ]
(a) Eigenvalues of the given matrix
Let λ is the eigenvalues of the given matrix in such a way that
det ( A−¿ λI )=0 ¿
Now,
A−λI = [3 2 −1
3 8 −3
3 6 −1 ]− λ [1 0 0
0 1 0
0 0 1 ]
A−λI = [3 2 −1
3 8 −3
3 6 −1 ]− [ λ 0 0
0 λ 0
0 0 λ ]
A−λI = [3−λ 2 −1
3 8− λ −3
3 6 −1−λ ]
Determinant of A−λI
| A−λI |=
|
3−λ 2 −1
3 8−λ −3
3 6 −1−λ |
¿ ( 3−λ ) ( λ2 −7 λ+10 ) −2 ( 6−3 λ )−1 .3 ( λ−2 )
¿ ( 3−λ ) ( λ−2 ) ( λ−5 )−12+6 λ−3 λ+6
¿ ( 3−λ ) ( λ−2 ) ( λ−5 )+3 λ−6
¿ ( 3−λ ) ( λ−2 ) ( λ−5 )+3 ( λ−2 )
32

¿ ( λ−2 ) { ( 3− λ ) ( λ−5 ) + 3 }
¿ ( λ−2 ) { 3 λ−15−λ2+5 λ+ 3 }
¿ ( λ−2 ) {−λ2 +8 λ−12 }
¿− ( λ−2 ) { λ2−8 λ +12 }
¿− ( λ−2 ) ( λ−6 ) ( λ−2 )
det ( A−¿ λI )=− ( λ−2 ) ( λ−6 ) ( λ−2 ) ¿
Now,
det ( A−¿ λI )=0 ¿
− ( λ−2 ) ( λ−6 ) ( λ−2 ) =0
λ=2 ,2 , 6
Therefore, the eigenvalues of matrix A is 2 , 2, 6 .
(b) Eigenvectors for each of the eigenvalue
Eigenvectors
Let the eigenvector is ϑ, such that ( A−λ I ) ϑ =0 for each of the corresponding eigenvalue.
Now,
For eigenvalue λ=2 ,
( A−2 I ) ϑ=0
33

( [3 2 −1
3 8 −3
3 6 −1 ]−2 [1 0 0
0 1 0
0 0 1 ]) [ x
y
z ]= [0
0
0 ]
[1 2 −1
3 6 −3
3 6 −3 ][ x
y
z ]= [0
0
0 ]
Now reduce the matrix into row echelon form as
[a … . b
0 … . …
0 0 c ]
[ 1 2 −1
3 6 −3
3 6 −3 ] : [ 3 6 −3
0 0 0
0 0 0 ]
Now reduce the matrix into reduced row echelon form as
[1 … . b
0 … . …
0 0 1 ]
[3 6 −3
0 0 0
0 0 0 ]: [1 2 −1
0 0 0
0 0 0 ]
Hence,
[1 2 −1
0 0 0
0 0 0 ][ x
y
z ]= [0
0
0 ]
x +2 y −z=0 and x=−2 y + z
z=0
34

Now,
[ x
y
z ]= [−2 y +z
y
z ] or ¿ [
−2 y
y
0 ] + [ z
0
z ]
Where, y∧z ≠ 0
Let y=1∧z=1
hence ,
Eigenvector for λ=2 is [−2
1
0 ], [ 1
0
1 ]
For eigenvalue λ=6 ,
( A−2 I ) ϑ=0
( [3 2 −1
3 8 −3
3 6 −1 ]−6 [1 0 0
0 1 0
0 0 1 ] ) [ x
y
z ]= [0
0
0 ]
[−3 2 −1
3 2 −3
3 6 −7 ][ x
y
z ]= [0
0
0 ]
Now reduce the matrix into row echelon form as
[a … . b
0 … . …
0 0 c ]
[ −3 2 −1
3 2 −3
3 6 −7 ] : [ −3 2 −1
0 8 −8
0 0 0 ] 35

Now reduce the matrix into reduced row echelon form as
[1 … . b
0 … . …
0 0 1 ]
[ −3 2 −1
0 8 −8
0 0 0 ] :
[ 1 0 −1
3
0 1 −1
0 0 0 ]
Hence,
[ 1 0 −1
3
0 1 −1
0 0 0 ] [ x
y
z ]= [ 0
0
0 ]
x− ( 1
3 ) z =0 ,∨x= 1
3 z
y−z =0 ¿ y=z
Now,
[ x
y
z ]=
[ 1
3 z
z
z ]
Where, z ≠ 0
Let z=1
hence ,
36

Eigenvector for λ=6 is
[ 1
3
1
1 ]
Therefore, the eigenvectors are [ −2
1
0 ] , [ 1
0
1 ] ,
[ 1
3
1
1 ]
(c) Condition B=P−1 A P
Matrix P=?
Matrix A = [3 2 −1
3 8 −3
3 6 −1 ]
P matrix would be determined based on the eigenvectors of matrix A.
P=
[−2 1 1
3
1 0 1
0 1 1 ]
Now for Matrix B:
P−1=
[ −2 1 1
3
1 0 1
0 1 1 ]
−1
¿
[−2 1 1
3
1 0 1
0 1 1
⋮
1 0 0
0 1 0
0 0 1 ]
Now reduce matrix to row echelon form
37

[a … . b
0 … . …
0 0 c ]
R 2← R 2+( 1
2 ) R 1
[−2 1 1
3
0 1
2
7
6
0 1 1
⋮
1 0 0
1
2 1 0
0 0 1 ]
R 2↔ R 3
[ −2 1 1
3
0 1 1
0 1
2
7
6
⋮
1 0 0
0 0 1
1
2 1 0 ]
R 3 ← R 3− ( 1
2 )R 2
[ −2 1 1
3
0 1 1
0 0 2
3
⋮
1 0 0
0 0 1
1
2 1 −1
2 ]
Reduce the matrix to reduced row echelon form
[1 … . b
0 … . …
0 0 1 ]
38

[ 1 0 0
0 1 0
0 0 1
:
−3
4
−1
2
3
4
−3
4
−3
2
7
4
3
4
3
2
−3
4
]
P−1=
[−3
4
−1
2
3
4
−3
4
−3
2
7
4
3
4
3
2
−3
4
]
Now,
B=P−1 A P
¿
[ −3
4
−1
2
3
4
−3
4
−3
2
7
4
3
4
3
2
−3
4
] . [ 3 2 −1
3 8 −3
3 6 −1 ] .
[ −2 1 1
3
1 0 1
0 1 1 ]
¿
[−3
2 −1 3
2
−3
2 −3 7
2
9
2 9 −9
2 ].
[−2 1 1
3
1 0 1
0 1 1 ]
B= [ 2 0 0
0 2 0
0 0 6 ]
39

Hence, the matrix B is [2 0 0
0 2 0
0 0 6 ] .
To determine whether the computed matrix is correct on not based on the eigenvalues
determined in part (a).
If det (B) = multiplication of the eigenvalues
Now,
Debt ( B ) =
|
2 0 0
0 2 0
0 0 6|=2 ( 12 ) −0 ( 0 ) +0 ( 0 ) =24
Multiplication of the eigenvalues (From part a) = 2 * 2 * 6 = 24
It is apparent that both the sides are same and therefore, the computed matrix B is correct.
40

41