Consider the following system of linear equations: (x2) + 2(x3) = 7 (x1) -2(x2)
ID: 1891470 • 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
1) Solve the system of equations by using the method of Gauss-Jordan elimination.
Final answer should be the reduced row echelon form (rref) of the augmented matrix.
2) Write down the system of equations that is equivalent to the original system and that
corresponds the reduced-row echelon form obtained in part (1)
3) Solve the original system of equations by reading off the solution of its equivalent system in part (2)
4) 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
1)
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]
2)
so we have
x1 = 2/5
x2 = 5
x3 = 1