Linear Algebra Exam Solution: Problems on Systems, Surfaces, and Bases

This document presents a comprehensive solution to a Linear Algebra exam, addressing a variety of problems. The solution includes detailed steps for finding the best solution using the least squares method for a system of equations, solving an initial value problem involving a system of differential equations, identifying the type of a surface given its equation and determining the points on the surface closest to the origin. Furthermore, the solution covers topics such as orthonormal bases, linear subspaces, and diagonalizability of matrices, providing a complete guide for understanding and solving these types of linear algebra problems. The exam also explores concepts related to eigenvalues and eigenvectors, providing a strong foundation for mastering Linear Algebra concepts.

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 ]