Use Cramer's rule,if possible, to solve the system of linear equations. Algebra!!!?

Use Cramer's rule,if possible, to solve the system of linear equations.

x- y+2z=4

5x+ z=3

x+3y+ z=9

Question 1 answers

(3, 2, 0)

(2, 0, 3)

(3, 0, 2)

(0, 2, 3)

1 Answer

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  • 1 decade ago
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    The answer is (0, 2, 3).

    Here’s how you get it, using Cramer’s rule: you take 4 determinants.

    First, you evaluate the determinant of the co-efficient matrix:

    | +1 –1 +2 |

    | +5 +0 +1 |

    | +1 +3 +1 |.

    This we call D, and D = 31.

    Next, you replace the first column with the constants, 4, 3, and 9, and evaluate the determinant of that matrix:

    | +4 –1 +2 |

    | +3 +0 +1 |

    | +9 +3 +1 |.

    This we call D_x, and we find D_x = 0.

    Next, you replace the second column with the constants, 4, 3, and 9, and evaluate the determinant of that matrix:

    | +1 +4 +2 |

    | +5 +3 +1 |

    | +1 +9 +1 |.

    This we call D_y, and we find D_y = 62.

    Next, you replace the third column with the constants, 4, 3, and 9, and evaluate the determinant of that matrix:

    | +1 –1 +4 |

    | +5 +0 +3 |

    | +1 +3 +9 |.

    This we call D_z, and we find D_z = 93.

    Finally, we divide D into D_x, D_y, and D_z, to get x, y, and z, respectively:

    x = D_x / D = 0 / 31 = 0.

    y = D_y / D = 62 / 31 = 2.

    z = D_z / D = 93 / 31 = 3.

    * * *

    In case you are not familiar with the evaluation of 3 X 3 determinants, I will show you an easy method:

    | a b c |

    | d e f |

    | g h i |

    is evaluated as follows: D = aei + bfg + cdh – ceg – afh – bdi.

    In the case of the first determinant,

    | +1 –1 +2 |

    | +5 +0 +1 |

    | +1 +3 +1 |,

    you can write it like this:

    +1 –1 +2 +1 –1

    +5 +0 +1 +5 +0

    +1 +3 +1 +1 +3,

    repeating the first two columns, after the third column.

    Then, you multiply the three forward diagonals:

    +1 X +0 X +1 = 0; –1 X +1 X +1 = –1; +2 X +5 X +3 = 30.

    Add these three products and get 0 –1 + 30 = 29.

    Next, take the products of the three backward diagonals:

    +2 X +0 X +1 = 0; +1 X +1 X +3 = 3; –1 X +5 X +1 = –5.

    Add these three products and get 0 + 3 – 5 = –2.

    Subtract the second sum from the first sum and get 29 – (-2) = 31.

    This is your first determinant.

    You do all four determinants by the same method.

    I hope that this answer will be of help to you.

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