X1+2x2-3x3=8
x1+3x1-5x3=11
2x1+5x2-8x3=8
Determine:
1. The rank of the coefficient matrix
2. The adjacent of the coefficient matric
3. The inverse of the coefficient matric
4. The general solution
To determine the rank of the coefficient matrix, we need to write down the augmented matrix and perform row operations to reduce it to row-echelon form. The coefficient matrix for this system of equations is:
| 1 2 -3 |
| 1 3 -5 |
| 2 5 -8 |
Adding -1 times the first row to the second row, and adding -2 times the first row to the third row, we get:
| 1 2 -3 |
| 0 1 2 |
| 0 1 -2 |
Adding -1 times the second row to the third row, we obtain:
| 1 2 -3 |
| 0 1 2 |
| 0 0 -4 |
This is in row-echelon form. The rank of the coefficient matrix is 3, since there are 3 non-zero rows.
To find the adjugate of the coefficient matrix, we need to find the determinant of each 2x2 minor matrix and then arrange them into a matrix and take its transpose. The adjugate matrix is used to find the inverse of the coefficient matrix.
The adjugate of the coefficient matrix is:
| 8 -6 -3 |
| 11 -8 -3 |
| 8 -5 -2 |
To find the inverse of the coefficient matrix, we can use the formula:
inverse(A) = (1/det(A)) * adjugate(A)
where det(A) is the determinant of A.
The determinant, det(A), can be found by taking the sum of the products of the elements of each row with their cofactors. Starting with the first row, the cofactors are: +1, -1, and +2. Therefore:
det(A) = 1(1) + 2(-1) + (-3)(2) = 1 - 2 - 6 = -7
Then, the inverse of the coefficient matrix is:
inverse(A) = (1/-7) * adjugate(A) = (-1/7) *
| 8 -6 -3 |
| 11 -8 -3 |
| 8 -5 -2 |
Simplifying this, we get:
inverse(A) =
| -8/7 6/7 3/7 |
| -11/7 8/7 3/7 |
| -8/7 5/7 2/7 |
To find the general solution, we can write the system of equations in matrix form and solve for the variables.
The matrix equation is:
Ax = b
where A is the coefficient matrix, x is the vector of variables, and b is the vector of constants.
The system of equations can be written as:
| 1 2 -3 | |x1| | 8 |
| 1 3 -5 | * |x2| = | 11 |
| 2 5 -8 | |x3| | 8 |
We can use the inverse of A to solve for x:
x = inverse(A) * b
Substituting the values, we get:
| x1 | | -8/7 6/7 3/7 | | 8 | | x1 | | -8/7 + 6/7 - 3/7 |
| x2 | = | -11/7 8/7 3/7 | * | 11 | = | x2 | = | -11/7 + 16/7 - 3/7 |
| x3 | | -8/7 5/7 2/7 | | 8 | | x3 | | -8/7 + 5/7 - 2/7 |
x1 = -1/7
x2 = 3/7
x3 = 1/7
Therefore, the general solution to the system of equations is:
x1 = -1/7 + (1/7)n
x2 = 3/7
x3 = 1/7
where n is any integer.
To determine the rank of the coefficient matrix, we will first write down the augmented matrix of the given system of equations:
```
| 1 2 -3 | 8 |
| 1 3 -5 | 11 |
| 2 5 -8 | 8 |
```
1. The rank of the coefficient matrix is the number of linearly independent rows in the augmented matrix. To determine this, we will perform row operations to bring the augmented matrix into row-echelon form:
```
| 1 3 -5 | 11 |
| 0 -1 2 | -3 |
| 0 0 0 | 0 |
```
The row-echelon form shows that there are two linearly independent rows. Therefore, the rank of the coefficient matrix is 2.
2. The adjacent of the coefficient matrix is obtained by removing the last column from the augmented matrix:
```
| 1 2 -3 |
| 1 3 -5 |
| 2 5 -8 |
```
3. To find the inverse of the coefficient matrix, we will write down the augmented identity matrix and perform row operations until the coefficient matrix is transformed into the identity matrix or the left part of the augmented matrix becomes an identity matrix:
```
| 1 2 -3 | 1 0 0 |
| 1 3 -5 | 0 1 0 |
| 2 5 -8 | 0 0 1 |
```
After performing row operations, the left part of the augmented matrix becomes the identity matrix, and the right part gives us the inverse of the coefficient matrix:
```
| 1 0 0 |
|-1 1 0 |
|-2 -1 1 |
```
Therefore, the inverse of the coefficient matrix is:
```
| 1 0 0 |
|-1 1 0 |
|-2 -1 1 |
```
4. To find the general solution, we will express each variable in terms of the free variables (if any). In this case, since the rank of the matrix is 2 and there are 3 variables, there is one free variable.
Let's assume x3 = t, where t is a free variable. Then, we can express x1 and x2 in terms of t using the equations:
From the first row: x1 + 2x2 - 3x3 = 8
x1 + 2x2 - 3t = 8
x1 = 8 - 2x2 + 3t
From the second row: x1 + 3x1 - 5x3 = 11
x1 + 3x1 - 5t = 11
4x1 - 5t = 11
4(8 - 2x2 + 3t) - 5t = 11
32 - 8x2 + 12t - 5t = 11
-8x2 + 7t = -21
x2 = (7t - 21)/8
Therefore, the general solution to the system of equations is:
x1 = 8 - 2x2 + 3t
x2 = (7t - 21)/8
x3 = t
where t is a free parameter.