- #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.