X1+2x2-3x3=8

Question

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

Answer 1

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.