How can I determine the free variable is z and t in this sou

How can I determine the free variable is z and t in this soulution, I just confused in here. Can experts give me hints that how can I determine after Guassian elimination?

2. Let W be the solution space of the following homogeneous system: 3x+11y-4z +10s -9t 0 Find the dimension and a basis for W. Solution: Reduce the system to echelon form: 3x+11y-4z +10s -9t 0 In echelon form, the system has 2 free variables, z and t; hence dim(W) 2. A basis tun,u2 for W may be obtained as follows: (1) Set z 1, t 30. Back substitution yields s 0, then y 1, and then x 5. Thus, u, [5 -1 1 0 0\'. (2) Set z 0, 1. Back substitution yields s 2, then y 5, and then x -2. Thus D-22 5 0 2 11

Solution

We have 3 equations in 5 variables viz. x,y,z, s and t. When we solve these equations, there will be 2 independent variables and 3 dependent variables i.e. 3 variables can be determined in terms of 2 other variables. The independent variables are such that the dependent variables can easily be expressed in terms of the independent variables. Let us attempt to solve the actual question to make things clear. The coefficient matrix of the given homogeneous linear system is A =

1

3

-2

5

-3

2

7

-3

7

-5

3

11

-4

10

-9

To solve the given linear equations, we will reduce A to its RREF by the following row operations:

Add -2 times the 1st row to the 2nd row

Add -3 times the 1st row to the 3rd row

Add -2 times the 2nd row to the 3rd row

Add 3 times the 3rd row to the 2nd row

Add -5 times the 3rd row to the 1st row

Add -3 times the 2nd row to the 1st row

Then the RREF of A is

1

0

-5

0

22

0

1

1

0

-5

0

0

0

1

-2

Thus, the given equations are equivalent to

x-5z+22t = 0…(1)

y+z-5t = 0…(2)

s-2t = 0…(3)

It may be observed now that all the three equations include t so that t will be one of the independent variables. Further, it is apparent from the 3^rd equations that s can be expressed in terms of t so that s will be one of the dependent variables.This leaves us with x,y and z. Now, both the 1^st and the 2^nd equation have z and t   so that x and y can be expressed in terms of z and t. This makes z , the 2^nd independent variable. The remaining 3 variables i.e. x,y,s ( dependent variables) can be expressed in terms of z and t(independent variables).

The solution is (x,y,z,s,t) = ( 5z-22t, -z+5t, z, 2t, t) = z(5,-1,1,0,0)+t(-22,5,0,2,1). There will be infinite solutions which can be obtained by assigning arbitrary values to z and t.

1	3	-2	5	-3
2	7	-3	7	-5
3	11	-4	10	-9