Homework SourseFind the set of solutions for the linear system 28 2x1x26x33
Find the set of solutions for the linear system.
28. 2x1-x2+6x3=-3 (This system has one equation.)
Each Linear system is not in echelon form, but can be put in echelon form by reordering the equation, Write the system in echelon form, and then find the set of solution.
32. (a) -2x2+x3=5x4=0
x1+3x2=2x3+2x4=1
(b) x2-4x3+3x4=2
x1-5x2-6x3+3x4=3
- 3x4=15
5x3-4x4=10
Find the set of solutions to the linear system
34.
x1-xz+2x3=-2
X2 3x3=1
0=1
Solution
28. We have 2x1-x2+6x3=-3. This is a single equation in 3 variables so that one variable can be expressed in terms of the other 2 variables. Thus, the linear system will have infinite solutions. Here, x2 =2x1+6x3+3 so that (x1,x2,x3) = (x1, 2x1+6x3+3,x3) = x1( 1,2,0)+x3( 0,6,1)+(0,3,0).The solutions can be determined by assigning arbitrary values to x1 and x3.
32. (a)The given linear system is
-2x2+x3-5x4 = 0
x1+3x2-2x3+2x4=1.
The matrix representation of this system is AX = b, where b = (0,1)Tand A =
0
-2
1
-5
1
3
-2
2
The augmented matrix of the given system of equations is B =
0
-2
1
-5
0
1
3
-2
2
1
To solve these equations, we will reduce B to its RREF as under:
Interchange the 1st row and the 2nd row
Multiply the 2nd row by -1/2
Add -3 times the 2nd row to the 1st row
Then the RREF of B is
1
0
-1/2
-11/2
1
0
1
-1/2
5/2
0
Therefore, the given system of equations is equivalent to
x1-x3/2-11x4/2=1 and x2-x3/2-5x4/2 = 0 so that x1 = x3/2+11x4/2+1 and x2 = x3/2+5x4/2. This is a system of 2 equations in 4 variables so that 2 variables can be expressed in terms of the other 2 variables. The solution is (x1,x2,x3,x4) = (x3/2+11x4/2 +1, x3/2+5x4/2,x3,x4) = (1,0,0,0)+x1(1/2, 1/2,1,0)+x4((11/2,5/2 ,0,1). The solutions can be determined by assigning arbitrary values to x3 and x4.
(b) The given linear system is
x2-4x3+3x4=2
x1-5x2-6x3+3x4=3
- 3x4=15
5x3-4x4=10
The matrix representation of this system is AX = b, where b = (2,3, 15,10)Tand A =
0
1
-4
3
1
-5
-6
3
0
0
0
-3
0
0
5
-4
The augmented matrix of the given system of equations is B =
0
1
-4
3
2
1
-5
-6
3
3
0
0
0
-3
15
0
0
5
-4
10
To solve these equations, we will reduce B to its RREF as under:
Interchange the 1st row and the 2nd row; Interchange the 3rd row and the 4th row
Multiply the 3rd row by 1/5; Multiply the 4th row by -1/3
Add 4/5 times the 4th row to the 3rd row; Add -3 times the 4th row to the 2nd row
Add -3 times the 4th row to the 1st row; Add 4 times the 3rd row to the 2nd row
Add 6 times the 3rd row to the 1st row; Add 5 times the 2nd row to the 1st row
Then the RREF of B is
1
0
0
0
51
0
1
0
0
9
0
0
1
0
-2
0
0
0
1
-5
Therefore, the given linear system is equivalent to x1= 51,x2 =9, x3= -2 and x4 = -5 Which is the unique solution to the given linear system.
34. The given system is inconsistent as we cannot have 0 = 1.
| 0 | -2 | 1 | -5 |
| 1 | 3 | -2 | 2 |
