SIT292 Linear Algebra Assignment 2 Solution: 2017, Deakin University

This document provides a detailed solution to SIT292 Linear Algebra Assignment 2 from 2017, covering key concepts such as matrix cofactors, orthogonal vectors, and the Gaussian elimination method. The solution includes step-by-step calculations and explanations for determining eigenvalues and eigenvectors for various matrices, along with verification steps. The assignment addresses concepts like matrix diagonalization and the conditions required for it. The solution also provides insights into matrix operations, systems of equations, and their consistency. This document is a valuable resource for students studying linear algebra, offering a comprehensive guide to solving complex problems and understanding the underlying principles.

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