The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+4 votes
547 views

Consider the following set of equations

$$x+2y=5\\
4x+8y=12\\
3x+6y+3z=15$$

This set

  1. has unique solution

  2. has no solution

  3. has finite number of solutions

  4. has infinite number of solutions

 

asked in Linear Algebra by Veteran (67.5k points) | 547 views
A|B matrix will be given as -

$\begin{bmatrix} 1 &2 &0 &|5 \\ 4& 8 &0 & |12 \\ 3& 6 & 3 & |15 \end{bmatrix}$

After the row-operations -   R3->R3-3R1  and

                                            R2->4R1-R2

We obtain the matrix -

  $\begin{bmatrix} 1 &2 &0 &|5 \\ 0 & 0& 0& |8\\ 0& 0 & 0 & |0 \end{bmatrix}$

Now since, $Rank(A)\neq Rank(A|B)$

Hence, this is an inconsistent solution, that is, no solution.

2 Answers

+12 votes
Best answer
There are no solutions.

If we multiply 1st equation by 4, we get

4x + 8y = 20

But 2nd equation says

4x + 8y = 12

Clearly, there can not be any pair of (x,y), which satisfies both equations.
answered by Veteran (11k points)
selected by
I am getting rank of augmented matrix as 3 as well as rank of A =3 so according to this condition there must be a unique solution , although it shouldn't exist but then how can it contradict ?
Rank of A is not 3. Recheck your method.
+4 votes

The rank of augemented matrix(|AB|=3) and coefficient matrix|A|=2 is not same! hence there is no solution.


 

answered by Veteran (16.8k points)


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

29,017 questions
36,844 answers
91,383 comments
34,722 users