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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
in Linear Algebra by Veteran (52.1k points) | 1.8k views
+10
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 -

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

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

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

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.

 

0

..

0

The error is still there in GO hardcopy- 2019 

0
With this print mistake, if we solve the question we get Option A) has unique solution. :P

Please let me know if I'm wrong. :)

2 Answers

+30 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.

Correct Answer: $B$
by Boss (11.2k points)
edited by
0
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 ?
+2
Rank of A is not 3. Recheck your method.
+3 votes

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

by Loyal (9.6k points)
Answer:

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