2x1-x2-3x3=0 3x1+4x2+2x3=1 x1+5x2+x3= -3

Предыдущий
вопрос Следующий
вопрос

To solve this system of equations, we can use the method of substitution or matrix operations. Here, we will use the matrix operations method to find the solution.

The system of equations can be written in matrix form as:

| 2 -1 -3 | | x1 | | 0 |
| 3 4 2 | * | x2 | = | 1 |
| 1 5 1 | | x3 | | -3 |

Let's denote the coefficient matrix as A, the variables matrix as X, and the constants matrix as B. So, we have AX = B.

The next step is to find the inverse of matrix A.

| A | = 1 / det(A) adj(A)
det(A) =2(41 - 25) - (-1)(31 - 21) + (-3)(35 - 41) = 2

adj(A) = | 4 2 3 |
| -1 3 -1 |
| 15 -10 -7 |

| A |^-1 = 1/2 * | 0 2 -3 |
| 1 -3 1 |
| 15 20 14 |

Now, multiply the inverse of A with matrix B to get the solution matrix X:

| X | = | A |^-1 * | B |

| X | = 1/2 * | 0 2 -3 | | 0 |
| 1 -3 1 | | 1 |
| 15 20 14 | | -3 |

| X | = | 1 |
| -1 |
| 2 |

Therefore, the solution to the system of equations is:
x1 = 1
x2 = -1
x3 = 2