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Systems of Linear Equations
Examples
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
Next Step:
Replace R2 with 2R1-3R2
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
Replace R2 with 2R1-3R2
2870
5131R2R
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
Replace R2 with 2R1-3R2
2870
5131R2R
Next Step:
Divide R2 by -7
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
Replace R2 with 2R1-3R2
2870
5131R2R
410
5131R2R
Divide R2 by -7
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
Replace R2 with 2R1-3R2
2870
5131R2R
410
5131R2R
Divide R2 by -7
Next Step:
Replace R1 with R1-R2
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
Replace R2 with 2R1-3R2
2870
5131R2R
410
5131R2R
Divide R2 by -7
Replace R1 with R1-R2
410
9031R2R
Next Step:
Divide R1 by 3
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
You already know how to solve this:
6y3x2)2E(
5yx3)1E(
Eliminate x by subtracting
2(E1)-3(E2)
28y7 Solve for y
4y Substitute this value into (E1)
5)4(x3 Solve for x
3x
9x3
5)4(x3
a) Here is the Gaussian elimination version:
6y3x2
5yx3
632
5131R2R
Replace R2 with 2R1-3R2
2870
5131R2R
410
5131R2R
Divide R2 by -7
Replace R1 with R1-R2
410
9031R2R Divide R1 by 3
410
3011R2R
The solution is right there in the matrix: x=-3, y=4
So the solution is x=-3, y=-4
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
10
01
32
131R2R
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
32
01
70
132R31R22R
10
01
32
13 1R2R
*1R2R
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
32
01
70
132R31R22R
10
01
32
13 1R2R
*1R2R
73
72
73
79
1R2R
*
73
72
1R2R7
1*
10
032R1R1R
01
10
132R2R
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
32
01
70
132R31R22R
10
01
32
13 1R2R
*1R2R
73
72
73
79
1R2R
*
73
72
1R2R7
1*
10
032R1R1R
01
10
132R2R
73
72
71
73
1R2R3
1*
10
011R1R
This is the inverse matrix
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
32
01
70
132R31R22R
10
01
32
13 1R2R
*1R2R
73
72
73
79
1R2R
*
73
72
1R2R7
1*
10
032R1R1R
01
10
132R2R
73
72
71
73
1R2R3
1*
10
011R1R
This is the inverse matrix
73
72
71
73
1A
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
32
01
70
132R31R22R
10
01
32
13 1R2R
*1R2R
73
72
73
79
1R2R
*
73
72
1R2R7
1*
10
032R1R1R
01
10
132R2R
73
72
71
73
1R2R3
1*
10
011R1R
This is the inverse matrix
73
72
71
73
1A
Now that we have the inverse, we can multiply on both sides of the original equation:
bAx
bAxAA1
1
Identity
1
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
32
01
70
132R31R22R
10
01
32
13 1R2R
*1R2R
73
72
73
79
1R2R
*
73
72
1R2R7
1*
10
032R1R1R
01
10
132R2R
73
72
71
73
1R2R3
1*
10
011R1R
This is the inverse matrix
73
72
71
73
1A
Now that we have the inverse, we can multiply on both sides of the original equation:
6
5x
bAx
bAxAA
73
72
71
73
1
1
Identity
1
where
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve the following system by
a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
6y3x2
5yx3
b) Here is the inverse method:
The system of equations is in the form
bxA
6
5b;
32
13A
We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right.
32
01
70
132R31R22R
10
01
32
13 1R2R
*1R2R
73
72
73
79
1R2R
*
73
72
1R2R7
1*
10
032R1R1R
01
10
132R2R
73
72
71
73
1R2R3
1*
10
011R1R
This is the inverse matrix
73
72
71
73
1A
Now that we have the inverse, we can multiply on both sides of the original equation:
where
This is the same answer we got before: X=-3, y=4
*Note: You might be thinking that the inverse matrix way was a whole lot of work to get an answer that we just got faster using the other method. And you would be correct. However, it is a much more general method that will apply to more complicated problems in the future (that would not be so easily solved by the other methods).
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4
3x
6
5x
bAx
bAxAA
73
72
71
73
1
1
Identity
1
Solve for x and y:
1y3x5
8y2x
22y8x6
1y2x5
Solve for x and y:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x and y:
1y3x5
8y2x
39
8
130
211R52R2R
1
8
35
21 *
3
2
10
012R21R1R
3
8
10
212R2R *
131*
The solution is x= -2, y= 3
22y8x6
1y2x5
Solve for x and y:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x and y:
1y3x5
8y2x
39
8
130
211R52R2R
1
8
35
21 *
3
2
10
012R21R1R
3
8
10
212R2R *
131*
The solution is x= -2, y= 3
22y8x6
1y2x5
Solve for x and y:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
104
1
520
251R62R52R
22
1
86
25 *
2
5
10
052R21R1R
2
1
10
252R2R *
521*
2
1
10
011R1R
51*
The solution is x=1, y=2
Solve for x and y:
4y2x3
4y3x2
Solve for x and y:
0yx10
17y2x3
Solve for x and y:
38y4x3
4y2x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x and y:
4y2x3
4y3x2
20
4
50
321R32R22R
4
4
23
32 *
4
8
10
022R31R1R
4
4
10
322R2R *
51*
4
4
10
011R1R
21*
The solution is x=-4, y=4
Solve for x and y:
0yx10
17y2x3
Solve for x and y:
38y4x3
4y2x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x and y:
4y2x3
4y3x2
20
4
50
321R32R22R
4
4
23
32 *
4
8
10
022R31R1R
4
4
10
322R2R *
51*
4
4
10
011R1R
21*
The solution is x=-4, y=4
Solve for x and y:
0yx10
17y2x3
170
17
170
231R102R32R
0
17
110
23 *
10
3
10
032R21R*1R
10
17
10
232R2R
171*
10
1
10
011R1R
31* The solution is x=1, y=-10
Solve for x and y:
38y4x3
4y2x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x and y:
4y2x3
4y3x2
20
4
50
321R32R22R
4
4
23
32 *
4
8
10
022R31R1R
4
4
10
322R2R *
51*
4
4
10
011R1R
21*
The solution is x=-4, y=4
Solve for x and y:
0yx10
17y2x3
170
17
170
231R102R32R
0
17
110
23 *
10
3
10
032R21R*1R
10
17
10
232R2R
171*
10
1
10
011R1R
31* The solution is x=1, y=-10
Solve for x and y:
38y4x3
4y2x
50
4
100
211R32R2R
38
4
43
21 *
5
4
10
212R*2R
101
5
6
10
012R21R*1R
5
6
10
011R*1R
The solution is x=6, y=5
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
1
1
1
437
215
349
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
16
14
1
1510
3110
349
1
1
1
437
215
3491R52R9*2R1R73R9*3R
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
162
14
1
16200
3110
349
2R3R11*3R
16
14
1
1510
3110
349
1
1
1
437
215
3491R52R9*2R1R73R9*3R
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
1
14
1
100
3110
349
3R*3R
162
14
1
16200
3110
349
2R3R11*3R
16
14
1
1510
3110
349
1
1
1
437
215
349
16211R52R9*2R
1R73R9*3R
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
1
14
1
100
3110
349
3R*3R
162
14
1
16200
3110
349
2R3R11*3R
16
14
1
1510
3110
349
1
1
1
437
215
349
16211R52R9*2R
1R73R9*3R
1
11
1
100
0110
349
3R32R*2R
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
1
14
1
100
3110
349
3R*3R
162
14
1
16200
3110
349
2R3R11*3R
16
14
1
1510
3110
349
1
1
1
437
215
349
16211R52R9*2R
1R73R9*3R
1
1
1
100
010
349
2R*2R
1
11
1
100
0110
349
3R32R*2R111
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
1
14
1
100
3110
349
3R*3R
162
14
1
16200
3110
349
2R3R11*3R
16
14
1
1510
3110
349
1
1
1
437
215
349
16211R52R9*2R
1R73R9*3R
1
1
1
100
010
349
2R*2R
1
11
1
100
0110
349
3R32R*2R111
1
1
0
100
010
009
3R32R41R*1R
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
1
14
1
100
3110
349
3R*3R
162
14
1
16200
3110
349
2R3R11*3R
16
14
1
1510
3110
349
1
1
1
437
215
349
16211R52R9*2R
1R73R9*3R
1
1
1
100
010
349
2R*2R
1
11
1
100
0110
349
3R32R*2R111
1
1
0
100
010
001
1R*1R
1
1
0
100
010
009
3R32R41R*1R91
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Solve for x, y and z:
1z4y3x7
1z2yx5
1z3y4x9
1
14
1
100
3110
349
3R*3R
162
14
1
16200
3110
349
2R3R11*3R
16
14
1
1510
3110
349
1
1
1
437
215
349
16211R52R9*2R
1R73R9*3R
1
1
1
100
010
349
2R*2R
1
11
1
100
0110
349
3R32R*2R111
1
1
0
100
010
001
1R*1R
1
1
0
100
010
009
3R32R41R*1R91
The solution is x=0, y=-1, z=1
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
1010100
7743R3RR 21
*2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
1010100
7743R3RR 21
*2
1110
7743RR 210
1*2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
1010100
7743R3RR 21
*2
1110
7743RR 210
1*2
1110
3303R4RR 21
*1
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
1010100
7743R3RR 21
*2
1110
7743RR 210
1*2
1110
3303R4RR 21
*1
1110
1101RR 13
1*1
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
1010100
7743R3RR 21
*2
1110
7743RR 210
1*2
1110
3303R4RR 21
*1
1110
1101RR 13
1*1
We can interpret this final form of our matrix as follows:
1zy
1zx
x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns.
1zy
1zx
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
1010100
7743R3RR 21
*2
1110
7743RR 210
1*2
1110
3303R4RR 21
*1
1110
1101RR 13
1*1
We can interpret this final form of our matrix as follows:
1zy
1zx
x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns.
1zy
1zx
We can write out the solution by choosing z as our parameter, call it t:
t
1t
1t
x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1zy2x
7z7y4x3
Solve this system:
Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all).
1121
7743
Here is the augmented matrix for this system.
1010100
7743R3RR 21
*2
1110
7743RR 210
1*2
1110
3303R4RR 21
*1
1110
1101RR 13
1*1
We can interpret this final form of our matrix as follows:
1zy
1zx
x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns.
1zy
1zx
We can write out the solution by choosing z as our parameter, call it t:
t
1t
1t
x
If we separate into homogeneous and particular:
ph xx
0
1
1
t
1
1
1
x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
