Solve the simultaneous equation X-3y=2 and 2X+4y=-1 using matrix (inverse) method?

All_for_one

2 years ago

To solve the simultaneous equations X - 3y = 2 and 2X + 4y = -1 using the matrix (inverse) method, we can represent the equations in matrix form as follows:

[A] [X] = [B]

Where:
[A] is the coefficient matrix,
[X] is the column matrix of variables (X and Y),
[B] is the column matrix of constants.

In this case, we have:

[A] = | 1 -3 |
| 2 4 |

[X] = | X |
| Y |

[B] = | 2 |
| -1 |

To solve for [X], we can use the formula:

[X] = [A]⁻¹ [B]

First, we need to find the inverse of matrix [A]. The inverse of a 2x2 matrix [A] is given by:

[A]⁻¹ = (1/det[A]) [D]

Where det[A] is the determinant of matrix [A], and [D] is the adjugate of matrix [A].

det[A] = (1*4) - (-3*2) = 4 + 6 = 10

Now, let's find the adjugate [D]:

[D] = | 4 3 |
| -2 1 |

Now, we can calculate [A]⁻¹:

[A]⁻¹ = (1/10) [D] = (1/10) * | 4 3 |
| -2 1 |

[A]⁻¹ = | 4/10 3/10 |
| -2/10 1/10 |

[A]⁻¹ = | 2/5 3/10 |
| -1/5 1/10 |

Now, we can multiply [A]⁻¹ by [B] to find [X]:

[X] = [A]⁻¹ [B] = | 2/5 3/10 | * | 2 |
| -1/5 1/10 | | -1 |

[X] = | (2/5)*2 + (3/10)*(-1) |
| (-1/5)*2 + (1/10)*(-1) |

[X] = | (4/5) - (3/10) |
| (-2/5) - (1/10) |

[X] = | (8/10) - (3/10) |
| (-4/10) - (1/10) |

[X] = | (5/10) |
| (-5/10) |

[X] = | 1/2 |
| -1/2 |

So, the solution to the simultaneous equations is:
X = 1/2
Y = -1/2