Linear algebra study realm

State and prove the distance formula in between: i) a point and a line, ii) a point and a plane.

i) Suppose that you have a line X = tu + A and a point P. The area of a parallelogram is ||AP x u||. It height is the closest distance between P and u. Thus, ||AP x u||/||u|| = dist

ii) Suppose that you have a plan ax + by + cz = d and a point P(p1,p2,p3). The distance is the projection of the vector AP, where A(a1,a2,a3) is a point on the plane, onto the normal.
dist = ||Proj_n AP|| = |AP*n|/||n|| = |((p1 - a1, p2 - a2, p3 - a3)T * (a,b,c)T)| / (a^2 + b^2 + c^2)^(1/2) = |(a(p1 - a1) + b(p2 - a2) + c(p3 - a3))|/(a^2 + b^2 + c^2)^(1/2) = |(ap1 + bp2 + cp3 - aa1 - ba2 - ca3)|/(a^2 + b^2 + c^2)^(1/2) = |(ap1 + bp2 + cp3 - d)|/(a^2 + b^2 + c^2)^(1/2)

State the formula or describe how to find the distance between iii) two parallel lines, iv) two skew lines, v) two parallel planes.

Suppose that you have X1 = tu + A and X2 = sv + B, use projection of the AB onto u x v. It's for parallel or skew-lines.

Suppose that you have ax + by + cz = d1 and ax + by + cz = d2, the shortest distance is |d2 - d1|/ (a^2 + b^2 + c^2)^(1/2)

State Cramer’s Rule.

If AX=b, such as A=(a1|a2|a3) is a nxn invertible matrix, and X = (x,y,z)T and b are vectors, then the solution for X are:
x = det(b|a2|a3)/detA, y = det(a1|b|a3)/detA, z = det(a1|a2|b)/detA.

State the conditions which are equivalent to the statement: “The nxn matrix A is invertible”.

State the relationships between the rank of a matrix and the dimension of its row space, column space and null space.

For matrix A mxn

dim(rowA)=dim(colA)=rankA
dim(nullA)= n - rank A

Ok, I don't have questions anymore.

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Yep, you deserve my congratulations. Can you continue this job for our Physics exam.

How come you guys are so st*pid! These are simple questions, for maths God sake!

Can you do this question?

Find a linear system of two equations whose solution set has the form
(x1,x2,x3,x4)T = t1(1,2,3,4)T + t2(4,3,2,1)T, t1,t2 in R3
Then find the solutions of that system presenting them in vector form s1v1 + s2v2 and, finally,
show that vectors v1 and v2 belong to the set defined above.

Sure!

From the set (x1,x2,x3,x4)T = t1(1,2,3,4)T + t2(4,3,2,1)T, you get
x1 = t1 + 4t2,
x2 = 2t1 + 3t2,
x3 = 3t1 + 2t2, and
x4 = 4t1 + t2

Reduce the matrix

1 4 x1
2 3 x2
3 2 x3
4 1 x4

You get
x3 - 3x1 - 2(x2 - 2x1) = x1 - 2x2 + x3 = 0
x4 - 4x1 - 3(x2 - 2x1) = 2x1 - 3x2 + x4 = 0

Reduce the matrix

1 -2 1 0
2 -3 0 1

You get
(x1,x2,x3,x4)T = s1(3,2,1,0)T + s2(-2,-1,0,1)T = s1/2 (6,4,2,0)T + s2(-2,-1,0,1)T = (s1 - 2t1 + 2t1)/2 (6,4,2,0)T + (s2 - t1 + t1)(-2,-1,0,1)T = (s1 - 2t1)/2 (6,4,2,0)T + 2t1/2 (6,4,2,0)T + (s2 - t1) (-2,-1,0,1)T + t1(-2,-1,0,1)T = (s1 - 2t1)/2 (6,4,2,0)T + (s2 - 1t1) (-2,-1,0,1)T + t1(6,4,2,0)T + t1(-2,-1,0,1)T = (s1 - 2t1)/2 (6,4,2,0)T + (s2 - 1t1) (-2,-1,0,1)T + t1(4,3,2,1)T = (s1 - 2t1)/3 (9,6,3,0)T + (s2 - 1t1)/4 (-8,-4,0,4)T + t1(4,3,2,1)T
Assume that (s1 - 2t1)/3 = (s2 - 1t1)/4 = t2
(s1 - 2t1)/3 (9,6,3,0)T + (s2 - 1t1)/4 (-8,-4,0,4)T + t1(4,3,2,1)T = t2(1,2,3,4)T + t1(4,3,2,1)T

If two matrices A and B both commute with ((0,-1)T|(1,0)T) show that A and B commute themselves, that is AB=BA.

Then A((0,-1)T|(1,0)T) = ((0,-1)T|(1,0)T)A and B((0,-1)T|(1,0)T) = ((0,-1)T|(1,0)T)B.
Let A = ((a1,a2)T|(a3,a4)T)
((a1,a2)T|(a3,a4)T)((0,-1)T|(1,0)T) = ((0,-1)T|(1,0)T)((a1,a2)T|(a3,a4)T)
((-a2,-a4)T|(a1|a3)T) = ((a3,-a1)T|(a4|-a2)T)

This is consistent if and only if -a2=a3 and a1=a4

Thus define A = ((a1,a2)T|(-a2,a1)T) and B = ((b1,b2)T|(-b2,b1)T)
AB = ((a1,a2)T|(-a2,a1)T)((b1,b2)T|(-b2,b1)T) =
a1b1-a2b2 a1b2+a2b1
-a2b1-a1b2 -a2b2+a1b1

BA=((b1,b2)T|(-b2,b1)T)((a1,a2)T|(-a2,a1)T) =
b1a1+b2a2 b1a2+b2a1
-b1a2-b2a1 -b2a2+b1a1

Thus, AB=BA

Show that if matrix B commutes with a 2x2 matrix A then it must be of the form sA+tI for some scalars s and t.

If B = sA+tI, the AB = A(sA+tI) = sA^2 + tA = (sA+tI)A = BA
If B is not sA+tI; suppose that B = sA+tC, then AB = A(sA+tC) = sA^2 + tAC ≠ sA^2 + tCA = (sA+tI)A = BA

Show that matrices which commute with the matrix A((1,3)T|(2,4)T) have the form sA+tI (for some scalars s and t).

Let B = ((b1,b2)T|(b3,b4)T)
AB = ((1,3)T|(2,4)T)((b1,b2)T|(b3,b4)T)
You'll get the system:
3b2 - 2b3 = 0
2b1 + 3b2 - 2b4 = 0
3b1 + 3b3 - 3b4 = 0

Find the solution for b1,b2,b3,b4, you'll get
B = r1((-1,1)T|(2/3,0)T) + r2((1,0)T|(0,1)T) = r1((-1,1)T|(2/3,0)T) + (r2 - r1 + r1)((1,0)T|(0,1)T) = r1((-1,1)T|(2/3,0)T) + r1((1,0)T|(0,1)T) + (r2 - r1)((1,0)T|(0,1)T) = r1((0,1)T|(2/3,1)T) + (r2 - 2r1 +r1)((1,0)T|(0,1)T) = r1/3((0,3)T|(2,3)T) + (r2 - r1)((1,0)T|(0,1)T) = r1/3((0,3)T|(2,3)T) + (r2 - r1 - r1/3 + r1/3)((1,0)T|(0,1)T) = r1/3((0,3)T|(2,3)T) + r1/3((1,0)T|(0,1)T)+ (r2 - r1 - r1/3)((1,0)T|(0,1)T) = r1/3((1,3)T|(2,4)T) + (r2 - r1 - r1/3)((1,0)T|(0,1)T)
Let r1 = s and r2 - r1 - r1/3 = t
Thus, B = r1/3((1,3)T|(2,4)T) + (r2 - r1 - r1/3)((1,0)T|(0,1)T) = sA+tI

Show that, if A and B are square matrices such that A B A =I then both A and B are invertible and A B = B A.

If ABA = I;
then A(BA) = I, BA is the inverse of A
(AB)A = I, AB is the inverse of A
A can only have one inverse, thus AB=BA

Therefore,
A(AB) = I; (AA)B = I, AA is the inverse of B
(BA)A = I; B(AA) = I, the same AA is the inverse of B.