Question

Find whether the following equations are consistent then solve them2x+3y+4z=11;x+5y+7z=15;3x+11yj13z=25

Answer 1

Answer:

The equations are consistent and

The solution is unique and the solution is

x=2, y=-3, z=4

Step-by-step explanation:

The given equation are

2x+3y+4z=11 , x+5y+7z=15 , 3x+11y+13z=25

Now the co-efficient matrix of the given equation is,

| 2 3 4 |

| 1 5 7 |

| 3 11 13 |

Now the determinant of the matrix = 2×(65-77) -3×(13-21) +4×(11-15) =-24+24-16= -16

So, the determinant of the matrix in non zero, so the rank of the matrix = 3 and the matrix has unique solution

Now the augmented matrix is

| 2 3 4 11 |

| 1 5 7 15 |

| 3 11 13 25 |

Now applying row elementary operations on the augmented matrix we are trying to reduce it in a row echolon matrix

Now, R1÷2 ,| 1 3/2 2 11/2 |

| 1 5 7 15 |

| 3 11 13 25 |

Now, R2-R1, and R3-3R1,

We get, | 1 3/2 2 11/2 |

| 0 7/2 5 19/2 |

| 0 13/2 7 17/2 |

Now R2×2/7,

We get , | 1 3/2 2 11/2 |

| 0 1 10/7 19/7|

| 0 13/2 7 17/2|

Now R1-3/2R2,and ,R3-13/2R2, we get,

| 1 0 -1/7 10/7 |

| 0 1 10/7 19/7 |

| 0 0 -16/7 -64/7 |

Now R3× -7/16, we get,

| 1 0 -1/7 10/7 |

| 0 1 10/7 19/7 |

| 0 0 1 4 |

Now, it is clear that rank of the augmented matrix= 3 = rank of the co-efficient matrix, so, the equation are consistent,

So, x -z/7=10/7, y + 10/7 z = 19/7 , z = 4

So, y = 19/7 - 40/7 = -21/7 = -3

And x= 10/7 + 4/7 = 14/7 = 2

So, the solution is, x=2, y=-3, z=4