Matrices:- Range and null space

  • #1
349
15

Homework Statement


Question is uploaded
I have completed till part iii and obtained correct answers
i. 2
ii. Basis for R:- { ( 2 3 -1 ) , (1 4 2 ) }
Cartesian equation; 2x-y+z=0
iii. Basis for Null:- { ( -3 2 0 1 ) , (2 -3 1 0 ) }

2. The attempt at a solution
I have problem in last part. I have calculated the value of k to be -9 but I cant proceed further.
I have two major queries.

1. I understand the main concept is to use a particular solution and then add some "gradient" to get an equation which will generate solutions. However why do we use the range space to calculate the particular solution?

2. Why is the general equation of the form
x = particular solution + a(1st Basis for Null vector) +b (2nd basis of null vector ) + ( in this case nothing)......
Or more precisely why are the basis of null vector used as "gradients"?

I am pre-uni student so I will be very delighted if the answers doesnt contain very complicated language.
 

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Answers and Replies

  • #2
pasmith
Homework Helper
2,019
651

Homework Statement


Question is uploaded
I have completed till part iii and obtained correct answers
i. 2
ii. Basis for R:- { ( 2 3 -1 ) , (1 4 2 ) }
Cartesian equation; 2x-y+z=0
iii. Basis for Null:- { ( -3 2 0 1 ) , (2 -3 1 0 ) }

2. The attempt at a solution
I have problem in last part. I have calculated the value of k to be -9 but I cant proceed further.
I have two major queries.

1. I understand the main concept is to use a particular solution and then add some "gradient" to get an equation which will generate solutions. However why do we use the range space to calculate the particular solution?

The range space is everything you can possibly get; therefore [itex]b[/itex] must be in the range if there exists a [itex]v[/itex] such that [itex]Mv = b[/itex].

Here, not every vector in [itex]\mathbb{R}^4[/itex] is in the range; this limits the value of [itex]k[/itex]. Geometrically [itex](8, 7, k)[/itex] is a line parallel to the z-axis which doesn't pass through the origin, and R is a plane containing the origin which is not parallel to the z-axis; these must therefore intersect at a unique point.

2. Why is the general equation of the form
x = particular solution + a(1st Basis for Null vector) +b (2nd basis of null vector ) + ( in this case nothing)......
Or more precisely why are the basis of null vector used as "gradients"?

By definition, if [itex]u \in \ker M[/itex] then [itex]Mu = 0[/itex]. Thus if [itex]v[/itex] is such that [itex]Mv = b[/itex] then [itex]M(v + u) = Mv = b[/itex].
 
  • #3
349
15
Oh so technically we are just finding the coordinates of the intersection and then finding the value of k. After obtaining the matrix, we treat it as a position vector and then find the equation of a line which when substituted in the equation of plane gives us (8 7 -9).
I dont know whether my phrasing makes sense but the algebra in the last part of your answer made it all clear
 

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