Consider the following system of linear equations: (x2) + 2(x3) = 7 (x1) -2(x2)
ID: 1891527 • Letter: C
Question
Consider the following system of linear equations:(x2) + 2(x3) = 7
(x1) -2(x2) -6(x3) = -18
- (x1) -(x2) -2(x3) = -5
2(x1) -5(x2) -15(x3) = -46
The solution for this system is:
x1 = 2/5, x2 = 5, x3 = 1
a) Solve the original system of equations by reading off the solution of its equivalent system
b) What is the solution set of the homogeneous system of equations that is associated with given system. In other words, what is the solution set of the following homogeneous system of equations?
(x2) + 2(x3) = 0
(x1) -2(x2) -6(x3) = 0
- (x1) -(x2) -2(x3) = 0
2(x1) -5(x2) -15(x3) = 0
Explanation / Answer
(x2) + 2(x3) = 7
(x1) -2(x2) -6(x3) = -18
- (x1) -(x2) -2(x3) = -5
2(x1) -5(x2) -15(x3) = -46
a)
Gauss-Jordan elimination:
[ 0 1 2 | 7]
[ 1 - 2 - 6 | - 18]
[-1 -1 -2 | -5]
[ 2 -5 -15 | -46]
apply
R1 .<....> R2
[ 1 - 2 - 6 | - 18]
[ 0 1 2 | 7]
[-1 -1 -2 | -5]
[ 2 -5 -15 | -46]
apply R4 ...>R4 + 2R3
[ 1 - 2 - 6 | - 18]
[ 0 1 2 | 7]
[-1 -1 -2 | -5]
[ 0 -7 -19 | -56]
apply R4 ...>R4 + 7R2
[ 1 - 2 - 6 | - 18]
[ 0 1 2 | 7]
[-1 -1 -2 | -5]
[ 0 0 -5 | -7]
apply R3 .....>R3 -+R1
[ 1 - 2 - 6 | - 18]
[ 0 1 2 | 7]
[ 0 -3 -8 | -23]
[ 0 0 -5 | -7]
apply R3 ....> R3 + 3R2
[ 1 - 2 -6 | - 18]
[ 0 1 2 | 7]
[ 0 0 -2 | -2]
[ 0 0 -5 | -7]
apply R2 .... > R2 + R3
[ 1 - 2 - 6 | - 18]
[ 0 1 0 | 5]
[ 0 0 -2 | -2]
[ 0 0 -5 | -7]
apply R1 ..> R1 + 2R2
[ 1 0 - 6 | -8]
[ 0 1 0 | 5]
[ 0 0 -2 | -2]
[ 0 0 -5 | -7]
apply R3 ...> R3 /-2
[ 1 0 - 6 | -8]
[ 0 1 0 | 5]
[ 0 0 1 | 1]
[ 0 0 - 5 | -7]
apply R1....> 5R1 - 6R4
[ 5 0 0 | 2]
[ 0 1 0 | 5]
[ 0 0 1 | 1]
[ 0 0 - 5 |-7]
[ 5 0 0 | 2]
[ 0 1 0 | 5]
[ 0 0 1 | 1]
[ 0 0 0 |-2]
apply R1 ......> R1/5
[ 1 0 0 | 2/5]
[ 0 1 0 | 5]
[ 0 0 1 | 1]
[ 0 0 0 | -2]
so we have
x1 = 2/5
x2 = 5
x3 = 1
b)
(x2) + 2(x3) = 0
(x1) -2(x2) -6(x3) = 0
- (x1) -(x2) -2(x3) = 0
2(x1) -5(x2) -15(x3) = 0
we can write
[0 1 2 | 0]
[1 -2 -6 | 0]
[-1 -1 -2 | 0]
[2 -5 -15 | 0]
by performing row operation like above we get
[ 1 0 0 | 0]
[ 0 1 0 | 0]
[ 0 0 1 | 0]
[ 0 0 0 | 0]
so the solution is
X1 = 0
X2 = 0
X3 = 0