Solve the following systems of linear equations using the Gaussian elimination method.
x + 2y − 3z = 1
2x + 5y − 8z = 4
3x + 8y − 13z = 7.
x1 + 2x2 - 3x3 = 1
2x1 + 5x2 - 8x3 = 4
3x1 + 8x2 - 13x3 = 7
Rewrite the system in matrix form and solve it by Gaussian Elimination (Gauss-Jordan elimination)
1 2 -3 1
2 5 -8 4
3 8 -13 7
R2 - 2 R1 → R2 (multiply 1 row by 2 and subtract it from 2 row); R3 - 3 R1 → R3 (multiply 1 row by 3 and subtract it from 3 row)
1 2 -3 1
0 1 -2 2
0 2 -4 4
R1 - 2 R2 → R1 (multiply 2 row by 2 and subtract it from 1 row); R3 - 2 R2 → R3 (multiply 2 row by 2 and subtract it from 3 row)
1 0 1 -3
0 1 -2 2
0 0 0 0
Answer:
The system of equations has a solution set:
x1 + x3 = -3
x2 - 2x3 = 2
no
To solve the system of linear equations using the Gaussian elimination method, we will transform the system into an upper triangular form.
Step 1: Write the system of equations in matrix form:
[ 1 2 -3 | 1 ]
[ 2 5 -8 | 4 ]
[ 3 8 -13 | 7 ]
Step 2: Perform row operations to create zeros below the first entry in the first column. Multiply row 1 by -2 and add it to row 2. Multiply row 1 by -3 and add it to row 3.
[ 1 2 -3 | 1 ]
[ 0 1 -2 | 2 ]
[ 0 2 -4 | 4 ]
Step 3: Perform row operations to create zeros below the second entry in the second column. Multiply row 2 by -2 and add it to row 3.
[ 1 2 -3 | 1 ]
[ 0 1 -2 | 2 ]
[ 0 0 0 | 0 ]
Step 4: Solve for the variables by back substitution.
From the last row, we can see that 0 = 0. This indicates that we have infinitely many solutions. Let's solve for z and express x and y in terms of z:
z can take any value.
From the second row, we have:
y - 2z = 2
=> y = 2 + 2z
From the first row, we have:
x + 2y - 3z = 1
=> x + 2(2 + 2z) - 3z = 1
=> x + 4 + 4z - 3z = 1
=> x + z = -3
So, the solutions to the system of linear equations are given by:
x = -3 - z
y = 2 + 2z
z can take any value.
Therefore, the solutions to the system of linear equations are:
(x, y, z) = (-3 - z, 2 + 2z, z) where z is any real number.
To solve a system of linear equations using the Gaussian elimination method, we need to perform a series of row operations on the augmented matrix of the system until we reach an upper triangular form. Then, we can easily back-substitute and solve for the variables.
1. Write the augmented matrix for the system of equations:
| 1 2 -3 | 1 |
| 2 5 -8 | 4 |
| 3 8 -13 | 7 |
2. Perform row operations to create zeros below the first element in the first column:
(R2 - 2*R1) and (R3 - 3*R1)
| 1 2 -3 | 1 |
| 0 1 -2 | 2 |
| 0 2 -4 | 4 |
3. Next, create a zero in the third row, second column:
(R3 - 2*R2)
| 1 2 -3 | 1 |
| 0 1 -2 | 2 |
| 0 0 0 | 0 |
4. Now, we have an upper triangular form of the augmented matrix.
5. Perform back-substitution to solve for the variables:
From the third row, we can see that 0z = 0. This means z can take any value. We can assign it a parameter, let's say t.
Substituting z = t back into the second row, we get:
1y - 2t = 2
y - 2t = 2
From the second row, we can solve for y in terms of t:
y = 2 + 2t
Substituting y = 2 + 2t and z = t into the first row, we get:
x + 2(2 + 2t) - 3t = 1
x + 4 + 4t - 3t = 1
x + t = -3
Therefore, the solution to the system of equations is:
x = -3 - t, y = 2 + 2t, z = t
This means the system has infinitely many solutions, parameterized by t.