1.
We can find the equation of the best-fit line using linear regression. The general equation of a line is y = mx + b, where m is the slope of the line and b is the y-intercept.

To find the slope, we can use the formula:

m = (nΣ(xy) – ΣxΣy) / (nΣx^2 – (Σx)^2)

where n is the number of data points, Σ is the sum of, and x and y are the data points.

Plugging in the given data points:

n = 4
Σx = 1+2+3+4 = 10
Σy = -2-8-11-18 = -39
Σxy = 1*(-2)+2*(-8)+3*(-11)+4*(-18) = -123
Σx^2 = 1^2+2^2+3^2+4^2 = 30

m = (4*(-123) – 10*(-39)) / (4*30 – 10^2) = -5.1

To find the y-intercept, we can use the formula:

b = (Σy – mΣx) /n

Plugging in the values we calculated earlier:

b = (-39-(-5.1)*10)/4= 3

Therefore, the equation of the best-fit line is:

y = -5.1x+3

2.

(i)
To show that the set {u1, u2, …, un} is linearly independent, we need to prove that the only linear combination of these vectors that yields the zero vector is the trivial linear combination, where all coefficients are zero.

Suppose that there exist scalars c1, c2, …, cn, not all zero, such that:

c1u1 + c2u2 + … + cnun = 0

We can assume without loss of generality that c1 is non-zero. Then we can rewrite the equation above as:

u1 = (-c2/c1)u2 – … – (cn/c1)un

This implies that u1 is a linear combination of the other vectors in the set {u2, …, un}. But this contradicts the assumption that {u1, u2, …, un} are orthogonal vectors.

Recall that two non-zero vectors are orthogonal if and only if their dot product is zero. Therefore, for any two vectors ui and uj in the set {u1, u2, …, un}, where i≠j, we have:

ui ⋅ uj = 0

Multiplying both sides of the original equation by ui and taking the dot product with uj, where i≠j, we get:

(c1 ui ⋅ uj) + (c2 ui ⋅ uj) + … + (cn ui ⋅ uj) = 0

But since ui and uj are orthogonal, their dot product is zero. Thus, the only way for this equation to hold is if all the coefficients c1, c2, …, cn are zero. Therefore, the set {u1, u2, …, un} is linearly independent.

(ii)
Suppose that there exist two different representations of a vector v in terms of the set {u1, u2, …, un}:

v = α1u1 + α2u2 + … + αnun
v = β1u1 + β2u2 + … + βnun

We want to show that αi = βi for all i = 1, 2, …, n.

Taking the difference between the two representations, we have:

0 = (α1 – β1)*u1 + (α2 – β2)*u2 + … + (αn – βn)*un

Since the two representations are different, at least one coefficient must be non-zero. Without loss of generality, assume that α1 – β1 ≠ 0. Then we can solve for u1:

u1 = -[(α2 – β2)*u2 + … + (αn – βn)*un] / (α1 – β1)

This implies that u1 is a linear combination of the other vectors in the set {u2, u3, …, un}. But this contradicts the assumption that {u1, u2, …, un} are orthogonal vectors.

Recall that two non-zero vectors are orthogonal if and only if their dot product is zero. Therefore, for any two vectors ui and uj in the set {u1, u2, …, un}, where i≠j, we have:

ui ⋅ uj = 0

Multiplying both sides of the above equation by u1 and taking the dot product with uj, where i≠j, we get:

(α1 ui ⋅ uj – β1 ui ⋅ uj) + (α2 ui ⋅ uj – β2 ui ⋅ uj) + … + (αn ui ⋅ uj – βn ui ⋅ uj) = 0

But since ui and uj are orthogonal, their dot product is zero. Thus, the equation simplifies to:

(α1 – β1) ui ⋅ uj = 0

Since ui and uj are non-zero and orthogonal, their dot product is zero only if α1 – β1 = 0. Therefore, we have shown that αi = βi for all i = 1, 2, …, n. Thus, the representation of v in terms of {u1, u2, …, un} is unique.

(iii)
To find an expression of αj in terms of uj, we can take the dot product of both sides of the equation:

v ⋅ uj = (α1u1 + α2u2 + … + αn*un) ⋅ uj

By the distributive property of dot products, this simplifies to:

v ⋅ uj = α1*(u1 ⋅ uj) + α2*(u2 ⋅ uj) + … + αn*(un ⋅ uj)

But since the vectors {u1, u2, …, un} are orthogonal, their dot product is zero when i ≠ j, so we can simplify further:

v ⋅ uj = αj*(uj ⋅ uj)

Solving for αj, we get:

αj = (v ⋅ uj) / (uj ⋅ uj)

Therefore, the coefficient αj corresponding to vector uj in the representation of v as a linear combination of {u1, u2, …, un} is given by the above expression.

3.

x1=(1,2,3), x2=(2,3,4), x3=(2,5,7)

Folllowing the Gram-Schmidt process, we set v1 = x1 = (1,2,3).

Next, we have
v2 = x2-(<x2,x1>/||x1||^2)v1
= x2-(x2*x1/x1*x1)*v1
= (2, 3, 4)-[(1*2+2*3+3*4)/(1*1+2*2+3*3)]*(1,2,3)
= ((2, 3, 4)-(20/14)(1,2,3)
= (2, 3, 4)-(10/7)(1,2,3)
= (4/7,1/7,-2/7)

and
v3 = x3-(<x3,v1>/||v1||^2)v1-(<x3,v2>/||v2||^2)v2
= x3-[(2*1+5*2+7*3)/(1*1+2*2+3*3)](1,2,3)-[(2*(4/7)+5*(1/7)+7*(-2/7))/((4/7)*(4/7)+(1/7)*(1/7)+(-2/7)*(-2/7))](4/7,1/7,-2/7)
= (2,5,7)-(33/14)(1,2,3)-[(-1/7)/(21/49)](4/7,1/7,-2/7)
= (2,5,7)-(33/14,66/14,99/14)+(1/3)(4/7,1/7,-2/7)
= (-1/6, 5/14, -1/6)

We can verify that {(1,2,3),(4/7,1/7,-2/7),(-1/6, 5/14, -1/6)} forms an orthogonal basis for R^3.

Normalizing the vectors in the orthogonal basis, we obtain the orthonormal basis
{<1/√14, 2/√14, 3/√14>,<4/√21,1/√21,-2/√21>,<-7/√61,5/√61,-7/√61>}

4.
To find the eigenvectors of the symmetric matrix A = [13 -4; -4 7], we need to solve the characteristic equation:

det(A – λE) = 0, where I is the 2×2 identity matrix and λ is the eigenvalue we want to find.

det(A – λE) = det(13-λ -4;-4 7-λ) = (13 – λ)(7 – λ) – (-4)(-4) = 0
(λ-5)(λ-15)=0

so the eigenvalue is λ1 = 5, λ2 =15

For λ=5, (A-5E)=0

(13-5)x1-4×2=0
-4×1+(7-5)x2=0

Solving this system of equations, we get x1 = 1 and x2 = 2.

So the eigenvector corresponding to λ = 5 is v1 = [1 ; 2]

For λ=15, (A-15E)=0

(13-15)x1-4×2=0

-4×1+(7-15)x2=0

Solving this system of equations, we get x1 = -2 and x2 = 1.
So the eigenvector corresponding to λ = 15 is v1 = [-2 ; 1]

To verify that these eigenvectors are orthogonal, we can compute their dot product:

v1·v2 = (1)(-2) + (2)(1) = 0

Since the dot product is zero, the eigenvectors are orthogonal.

5.
The characteristic polynomial of a 2×2 matrix A is given by:

|A – λE| = det([a-λ c; c b-λ]) = (a – λ)(b – λ) – c^2

Expanding the determinant and simplifying, we get:

λ^2 – (a+b)λ + (ab – c^2) = 0

To show that the eigenvalues are real, we need to show that the solutions of this quadratic equation are real. We can use the discriminant to determine the nature of the roots:

D = (a+b)^2 – 4(ab – c^2) = a^2 + 2ab + b^2 – 4ab + 4c^2 = (a-b)^2 + 4c^2

Since a, b, and c are real numbers, the discriminant D is also real. Therefore, the roots of the quadratic equation are real if and only if the discriminant is non-negative:

D >= 0

Substituting the expression for D, we get:

(a-b)^2 + 4c^2 >= 0

Since the square of any real number is non-negative, the left-hand side of this inequality is always non-negative. Therefore, the discriminant is non-negative, and the roots of the characteristic polynomial are real.

Hence, the eigenvalues of a real symmetrical 2×2 matrix are always real.

6.
matrix form: [27 -5; -5 3][x;y]
A = [27 -5; -5 3]
det(A- λE) = det([27-λ -5; -5 3-λ]) = (27 – λ)(3 – λ) – 25 = λ^2 – 30λ + 56=(λ-2)(λ-28)

The roots of this quadratic equation are: λ1=2, λ2=28

These are the eigenvalues of the matrix A.

To find the eigenvectors, we need to solve the system of equations (A – λE)v = 0 for each eigenvalue.

For λ=2, (A-2E)=0

(27-2)x-5y=0
-5x+(3-2)y=0

Solving this system of equations, we get x1 = 1 and x2 = 5.
So the eigenvector corresponding to λ = 2 is v1 = [1 ; 5]

For λ=28, (A-28E)=0

(27-28)x-5y=0
-5x+(3-28)y=0

Solving this system of equations, we get x1 = 5 and x2 = -1
So the eigenvector corresponding to λ = 28is v1 = [5 ; -1]

A = PDP^(-1)

where
P = [v1 v2] = [1 5 ; 5 -1]
D = [λ1 0 ; 0 λ2] = [2 0 ; 0 28]

To find P^(-1), we can use the formula:

P^(-1) = (1/Δ) [d -b ; -c a], where Δ is the determinant of P and a, b, c, d are the cofactors of the elements of P:

Δ = |P| = (1)(-1) – (5)(5) = -26
a = -1, b = -5, c = -5, d = 1

Therefore,
P^(-1) = (1/-26) [1 5 ; 5 -1]

Now we can write the solution of the system of differential equations as:

[x ; y] = P [u ; v]

where [u ; v] is a new set of variables that diagonalize the system of differential equations:

[u ; v] = P^(-1) [x ; y]

Substituting the expressions for P and P^(-1), we get:

[u ; v] = (1/-26)[1 5 ; 5 -1][x ; y]
[u ; v] = [-y/26 + x/26 ; -5x/26 – y/26]

Therefore, the diagonalized system of differential equations is:

dx/dt = 2u = 2x/26 – 2y/26 = x/13 – y/13
dy/dt = 28v = -5x/26 + 5y/26 = -x/26 + y/26

The solution to this system of differential equations is:

x(t) = c1e^(t/13) + c2e^(-t/26)
y(t) = c1e^(t/13) – c2e^(-t/26)

where c1 and c2 are constants determined by the initial conditions.

7.
We have:

q(x) = 15x^2 – 6xy + 23y^2

Using the constraint ||x|| = 2, we can eliminate y in terms of x by solving for y in terms of x:

||x|| = 2
sqrt(x^2 + y^2) = 2
y^2 = 4 – x^2
y = sqrt(4 – x^2)

Substituting this into q(x), we get:

q(x) = 15x^2 – 6x(sqrt(4 – x^2)) + 23(4 – x^2)

Expanding and simplifying, we obtain:

q(x) = -8x^2 – 6x*sqrt(4 – x^2) + 92

To find the maximum and minimum values of q(x), we take the derivative with respect to x and set it to zero:

q'(x) = -(16*sqrt(2-x)*x*sqrt(x+2)-12*x^2+24)/(sqrt(2-x)*sqrt(x+2)) = 0

This equation does not have an exact solution, so we will use numerical methods to approximate the roots. Using a graphing calculator or a computer algebra system, we can find that the roots are approximately x ≈ -0.7 and x ≈ 1.8.

To determine the maximum and minimum values of q(x), we evaluate q(x) at the critical points and at the endpoints of the constraint ||x|| = 2:

q(-2) = 0
q(-0.7) ≈ 95.9
q(1.8) ≈ 56.6
q(2) = 0

Therefore, the maximum value of q(x) is approximately 95.9 and the minimum value is 0.