[FULL ACCESS] Linear Algebra and Matrix Operations

This assignment covers various topics in linear algebra and matrix operations. It includes four questions (Q1-Q4) that require students to perform calculations, solve systems of equations, find determinants, and apply transformation matrices. The solutions provided are detailed and step-by-step, making it easier for students to understand the concepts and techniques involved.

Q1
a
2 −2 −1
2 1 2
4 3 5
x
y
z
=
1
1
2
B
Getting 1 at the right hand conner
R1/2 = r1= (1 -1 -1/2 1/2)
1 −1 −1/2
2 1 2
4 3 5
1/2
1
2
Getting 0 on second row first column
-2r1+r2 = r2
(-2,2,1,-1)+(2,1,2,2)=(0,3,3,0)
1 −1 −1/2
0 3 3
4 3 5
1/2
0
2
Getting 0 on third row first column
-4r1+r3 = r3
(-4,4,2,-2)+(4,3,5,2)=(0,7,,0)
1 −1 −1/2
0 3 3
0 7 7
1/ 2
0
0
Getting 1 on second row second column
R3/3
1 −1 −1/2
0 1 1
0 7 7
1/ 2
0
0
Getting 0 on third row second column
-7r2+r3 = r3

(0,-7,-7,0)+(0,7,7,0 =(0,0,1,1)
1 −1 −1/2
0 1 1
0 0 0
1/ 2
0
0
Getting 1 on third row third column
r3 +1= r3
1 −1 −1/2
0 1 1
0 0 1
1/ 2
0
1
Getting 0 on second row third column
-r3 +r2= r2
1 −1 −1/2
0 1 0
0 0 1
1/ 2
−1
1
Getting 0 on first row third column
1/2r3 +r1= r1
1 −1 0
0 1 0
0 0 1
1
−1
1
Getting 0 on first row second column
R2 +r1= r1
1 0 0
0 1 0
0 0 1
0
−1
1
Therefore x=0, y=-1 and z=1
c.
Determinant of matrix A
2 −2 −1
2 1 2
4 3 5
= 2{(5*1-3*2) –2{(2*5-2*4)}+-1{2*3-4*1}

Determinant = -2
d.
the determinant is above is useful in finding the inverse of the matrix.
e.
c(3c-5)+2(4c-10)+1(-2)
3c2-5c+8c-20—2=0
3c2 +3c-22=0
C= -3.25 or 2.25
Q2
a.
1 1 2
2 1 2
4 3 5
x
y
z
=
2
1
2
b.
Determinant = -1+-2+4 = +1
We have 1 at the upper coner,
R1/2 = r1= (1 -1 -1/2 1/2)
1 1 2
2 1 2
4 3 5
2
1
2
Getting 0 on second row first column
-4r1+r2 = r2
1 1 2
0 −1 −2
4 3 5
2
−3
2
Getting 0 on third row first column
-4r1+r3 = r3

1 1 2
0 −1 −2
0 −1 −3
2
−3
−6
Getting 1 on second row second column
-R2/-1
1 1 2
0 1 2
0 −1 −3
2
3
−6
Getting 0 on third row second column
r2+r3 = r3
1 1 2
0 1 2
0 0 −1
2
3
−3
Getting 1 on third row third column
-r3 = r3
1 1 2
0 1 2
0 0 1
2
3
3
Getting 0 on second row third column
-2r3 +r2= r2
1 1 2
0 1 0
0 0 1
2
−3
3
Getting 0 on first row third column
-2r3 +r1= r1
1 1 0
0 1 0
0 0 1
−4
−3
3
Getting 0 on first row second column
-R2 +r1= r1

1 0 0
0 1 0
0 0 1
−1
−3
3
Therefore x=-1, y=-3 and z=3
c.solving using crammers rule
1 1 2
2 1 2
4 3 5
The determinant of the matrix will be 1(-1)-1(2)+2(2)= 1
Using constants to replace x-coefficients,
2 1 2
1 1 2
2 3 5
Dx =2(-1)-1(-1)+2(0) = -2+1+0 = -1
Using constants to replace y-coefficients
1 2 2
2 1 2
4 2 5
Dy= 1(1)-2(6)+4(2)= 1-12+8= -3
Using constants to replace z-coefficients,
1 1 2
2 1 1
4 3 2
Dz =1(-1)-2(-4)+4(-1)= -1+8-4=3
X= -1/1 = -1, y = -3/1= -3 and z= 3/1 = 3
d.
The determinant helps to get the inverse of the matrix
e.
Determinant = I(I2-C)-0(0*I-I*B)+0(0*C-I*B) = I(I2-C)
Q3

a.
Since the quadratic passes through (x1,y1), (x2,y2) and (x3 , y3) a matrix of
x 1 y 1
x 2 y 2
x 3 y 3
will be
formed. In addition, with the quadratic equation y=ax2 + bx + c, it means that the x values are
squared. And this leads to squaring the first value of x while the second accompanying will not
be squared
x 12 x y
x 12 x y
x 12 x y
b.
1 −1 1
4 2 1
25 5 1
=
1
−5
25
The upper coner is already having 1
Getting 0 on second row first column
-4r1+r2 = r2
1 −1 1
0 6 −13
25 5 1
1
−7
25
Getting 0 on third row first column
-25r1+r3 = r3
1 −1 1
0 6 −13
0 30 −24
1
−7
0
Getting 1 on second row second column
R2/6
1 −1 1
0 1 −13 /6
0 30 −24
1
−7 /6
0
Getting 0 on third row second column
-30r2+r3 = r3

1 −1 1
0 1 −13 /6
0 0 41
1
−7 /6
35
Getting 1 on third row third column
r3/41 = r3
1 −1 1
0 1 −13 /6
0 0 1
1
−7 /6
0.85
Getting 0 on second row third column
13/6r3 +r2= r2
1 −1 1
0 1 0
0 0 1
1
1.16
0.85
Getting 0 on first row third column
-r3 +r1= r1
1 −1 0
0 1 0
0 0 1
0.15
1.16
0.85
Getting 0 on first row second column
R2 +r1= r1
1 0 0
0 1 0
0 0 1
1.31
1.16
0.85
Therefore a=1.31, b=1.16 and c=0.85
C
In this equation, X2 represent on values enhancing the initial value f x which is x. therefore this
lead to (X1*1-X21*1)(X1*1- X3*1)(X2*1- X3*1) = (x1-x2)(x1-x3)(x2-x3)
d.
The equation must maintain the values of coordinates to ensure that the quadratic equation is
maintained.

e.
Q4
a.
T((x1x2x3)T = (x1-5x3, 3x1+x2)T
T(u+v) = T(u) +T(v) as well as T(αu) = αT(u)
T((x1x2x3)T =x1T(-5x3, 3+x2)T
Therefore
T(x2x3)T = (-5x3, 3+x2)T
Considering -5 and 3 as any other transformation factor, we conclude that T((x1x2x3)T is linear to
(x1-5x3, 3x1+x2)T since they can be factored in terms of x1x2 and x3
B
Transformation of a matrix
S : Rn_n = R
c.
v1 (1,1), v2 (1,-1), v3 (5,1), T(v1) = (-1,0) and T(v2) =(1,2)
t in v1 = (2,1) T in v2 is (0,-3) therefore T in v3 will be (-2,-7) leading tp T(v3) to be (3,-6)
d.
choosing a matrix
8 27
64 125
This matrix can be written in form of
4 9
16 25
2 3
4 5 multiplying these two matrices leads to the initial matrix of 8 27
64 125 and thus
showing a matrix can be expressed in form of A and I whose products is A3