Linear Algebra Solution 2022

Linear Algebra
The least square method of solving equations:
Given that x1 +x +x =1
x +x +x =1
x – x +x =2
2x –x –x =-1
Ax=b
[1 1 −1
1 1 1
1 −1 1
2 −1 −1 ]¿=
[ 1
1
2
−1 ]
ATA=ATb
[ 1 1 1
1 1 −1
−1 1 1
2
−1
−1 ] [1 1 −1
1 1 1
1 −1 1
2 −1 −1 ]¿= [ 1 1 1
1 1 −1
−1 1 1
2
−1
−1 ] [ 1
1
2
−1 ]
[ 7 −1 −1
−1 4 0
−1 0 4 ][ x 1
x 2
x 3 ]=
[2
1
3 ]
7x1 –x2-x3=2
-X1 +4x2=1
-x1+x3=3
Problem 2

X’=[ 3 4
2 5 ] and X(0)=[ 3
−1 ]
A= [ 3 4
2 5 ]
Then Solve for | A−ʎT |=0
But
| A−ʎT|=[ 3−ʎ 4
2 5−ʎ ]=0
This implies that;
(3-ʎ)(5-ʎ)-(2*4)=0
15-8ʎ +ʎ2-8=0
ʎ2-8ʎ +7=0
ʎ2-7ʎ-ʎ +7=0
Hence,
ʎ=7 or ʎ=1
Problem 3
The standard basis for |R3={
[1 0 0
0 1 0
0 0 1 ] } for the orthonormal basis.
But B={v1,v2,v3}
Which is the orthonormal basis for V
And X∈V
X=c1v1+c2v2 +c3v3 +…..+_ckvk

Vixi=3x12 +x22+3x23+x22+3X32-6x1x2+2x1x2 -6x2x3=1
=3x12 +2x22+6x23 -4x1x2 -6x2x3=1
And
ViX=Ci
Hence;
XB=[C 1
c 2
Ck ]=[ x 1
x 2
xk
x 1
x 2
xk ]
B={x1v2}
But v1=
[ 1
4
1
3 ] v2= [ 1
6
−1
2 ] v3=
[ 1
2
−1
2 ]
||v12||=1/16 +1/9
||v22||=1/36 +1/4
||V32||=1/4 +1/4
Problem 4.
The standard basis for |Rv={[1 0 0
0 1 0
0 0 1 ]}
Which is an orthornormal base
B ={V1,V2,v3|} is an orthonormal basis for U
Where;

x∈V
And U=2x1 +3X2+6x3=0
Hence
Vixi=2x1 +3x2 +6x3 which when projected to 0;
ViXi=Ci
[ 2 3 6 ] [ x 1
x 2
x 3 ]=[ 0 ]
XU=[ c 1
c 2
c 3 ]=[U 1
u 2
U 3
x 1
x 2
x 3 ]
B={X111U3)
U1*X1=0
U2*X2=0
U3*X3=0
Thus 0+0+0=0
Problem 5
A is diagnolizable if and only if A=I
Given that the square matrix (A-I)2=0
Ak=PDkP-1
For ʎ=1
A-ʎI= [ 3−1 1
1 3−1 ]=[ 2 1
1 2 ]

This implies that;
[ 2 2
0 0 ]
Hence
N(A-ʎI)=E(ʎ)=Span { [−2 ]
2 }
For ʎ=2
N(A-ʎI)=E(ʎ)={ [ −1
1 ]
[ xi
yi ]=PDkP-1(1
4 )
P= [−2 −1
2 1 ]
P-1= 1
−2−1 ([ 1 1
−2 −2 ]=
[−1
3
−1
3
2
3
2
3 ]
Problem 6
Problem a)
U as a linear subspace of Rn
Given that the vectors of Rn =u1,,,……um
Consider the standard basis for Rn =
[1 0 0
0 1 0
0 0 0
0
0
1 ]
B={u1,u2,u3,….un} Which is the orthonormal basis of U

Problem 6b
The diamension of U given that u1,…..um are linearly independent.
x→∈U
X→=c1u1 +C2u2 +c3U3 +……ckuk
uix→=c1uix1 +c2uix2 +c3uix3+……………+ciuixi +…..+ckulxk
v→iX=Ci
XB=[c 1
c 2
ck ]=[ x 1
x 2
xk
u1
u 2
uk ]
B=[x11u2}
U is a two dimension matrix
U1= [ 3
5
3
5 ] U2=
[ −4
5
3
5 ]
||u1||2= 9
25 + 16
25 =1
U1x2= −12
25 +12
25=0
=
[ 3
5
−3
5
3
5
2
5 ]