Equations solve by matrix method

(1) x+y+z=3
(2) x+2y+3z=4
(3) 2x+4y+9z=6

x+y+z=3

To solve a system of equations using the matrix method, we can represent the system of equations in matrix form.

First, let's rewrite the system of equations:

(1) x + y + z = 3
(2) x + 2y + 3z = 4
(3) 2x + 4y + 9z = 6

We can represent the coefficients of the variables (x, y, z) as a matrix, and the constants on the right-hand side of the equations as a separate matrix. The system can then be written as an augmented matrix [A|B], where A represents the coefficient matrix and B represents the constant matrix.

| 1 1 1 | | x | | 3 |
[ A | B ] = | 1 2 3 | | y | = | 4 |
| 2 4 9 | | z | | 6 |

Now, we can use row operations to simplify the augmented matrix until it is in its row-echelon form or reduced row-echelon form.

Let's perform the row operations:

1. Swap R1 and R2:
| 1 2 3 | | x | | 3 |
| 1 1 1 | | y | = | 4 |
| 2 4 9 | | z | | 6 |

2. Multiply R1 by -1 and add it to R2:
| 1 2 3 | | x | | 3 |
| 0 -1 -2 | | y | = | 1 |
| 2 4 9 | | z | | 6 |

3. Multiply R1 by -2 and add it to R3:
| 1 2 3 | | x | | 3 |
| 0 -1 -2 | | y | = | 1 |
| 0 0 3 | | z | | 0 |

4. Divide R3 by 3:
| 1 2 3 | | x | | 3 |
| 0 -1 -2 | | y | = | 1 |
| 0 0 1 | | z | | 0 |

5. Multiply R3 by 2 and add it to R2:
| 1 2 3 | | x | | 3 |
| 0 -1 0 | | y | = | 1 |
| 0 0 1 | | z | | 0 |

6. Multiply R2 by -1 and add it to R1:
| 1 1 3 | | x | | 2 |
| 0 -1 0 | | y | = | 1 |
| 0 0 1 | | z | | 0 |

7. Multiply R2 by -3 and add it to R1:
| 1 0 3 | | x | | -1 |
| 0 -1 0 | | y | = | 1 |
| 0 0 1 | | z | | 0 |

8. Multiply R3 by -3 and add it to R1:
| 1 0 0 | | x | | -1 |
| 0 -1 0 | | y | = | 1 |
| 0 0 1 | | z | | 0 |

Our augmented matrix is now in row-echelon form. From this form, we can read the values of x, y, and z directly:

x = -1
y = 1
z = 0

So, the solution to the system of equations is x = -1, y = 1, and z = 0.