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+14 votes

Consider the following set of equations


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 (59.9k points) | 1.5k 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


We obtain the matrix -

  $\begin{bmatrix} 1 &2 &0 &|5 \\ 0 & 0& 0& |8\\ 0& 0 & 3 & |0 \end{bmatrix}$

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

Hence, this is an inconsistent solution, that is, no solution.
@shweta how did you get 0000 oin the last row?
After performing row operation R3 -> R3 - 3R1
how 3 -0=0?
My mistake it should be 3. I edited it.

Formatting Error in PDF

There is an error in the formatting of this question in the soft copy of the GO PDF (screenshot attached for reference).

P.S: I am unable to add the screenshot. Please help me fix this.




3 Answers

+27 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 Boss (11.4k 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.
+8 votes

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


answered by Boss (17.3k points)
+3 votes

First transformed the matrix into echelon form and rest can seen from the pic.

answered by Loyal (9.2k points)

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