Linear algebra (trying to prepare for exam)

In summary, the conversation is discussing how to find a vector b such that the equation Ax = b has no solution for a given matrix A. The solution involves finding the reduced row echelon form of the matrix and using it to determine if the equation has a solution. It is also mentioned that the range of A is important in determining if a solution exists. Overall, the conversation highlights the importance of understanding the concepts behind solving equations rather than just applying formulas.
  • #1
frasifrasi
276
0
Ok, here we go.

The question asks,

for matrix A =

1 1 3
3 0 1
2 2 6

, find a vector b such that Ax(x is a vector) = b has no solution.


I found the rref(a) =

1 0 1/3
0 1 8/3
0 0 0

Now to, answer the question,

I gues the equation has no solution if one of the rows has 000 = a nonzero value.

So, is it right to say b is a with a nonzero value on the third row?
 
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  • #2
Find a basis for the range of A. If such a b exists then dim(range(A))<3. Find a vector that's not in the range.
 
  • #3
Is there an alternate way without having to work with range -- we didn't do this in class ?
 
  • #4
Well, you as you've said if b has a nonzero third component then you can't solve rref(A)x=b. Fine. But that doesn't mean you can't solve Ax=b. Now you have to 'undo' your row reduction operations on e.g. b=(0,0,1). Do the opposite row operations on that b to get a vector such that Ax=b doesn't have a solution.
 
  • #5
frasifrasi said:
Is there an alternate way without having to work with range -- we didn't do this in class ?
?? Surely they mentioned in class that Ax= b has a solution if and only if b is in the range of A! In fact, that is the definition of "range". You seem to be wanting to do what so many students unfortunately want- to apply a formula without having to understand what you are doing!

Saying Ax= b, here, is the same as saying x+ y+ 3z= bx, 3x+ z= by, and 2x+ 2y+ 6z = b3 and you have shown, by row reducing the matrix, that that corresponds to x+ (2/3)z= 2bx- (1/3)b[sub[y[/sub], y+ (1/3)z)= (1/3)b- a, 0= c- 2a. Looking at the last row we see that has a solution if and only if c- 2a= 0 or c= 2a. It does not have a solution is c is not 2a.

So, is it right to say b is a with a nonzero value on the third row?
No, it is not. For example, b= [1, 0, 2] has a nonzero value on the third row but Ax= b for this A and b is the same as x+ y+ 3z= 1, 3x+ z= 0, 2x+ 2y+ 6z= 2. From the second equation, z= -3x. Putting that into the first equation gives x+ y- 3x= -2x+ y= 1 or y= 2x+ 1. Putting that into the third equation, 2x+ 2(2x+ 1)+ 6(-x)= 2 which is true for all x- the xs cancel on the left.

It is true that an equation with the reduced matrix has no solution if there if the third component of b is non-zero. You should have carried the "b" values along with your row reduction to see what the right hand side should be for the original equation.
 
  • #6
"that that corresponds to x+ (2/3)z= 2bx- (1/3)b[sub[y[/sub], y+ (1/3)z)= (1/3)b- a, 0= c- 2a. Looking at the last row we see that has a solution if and only if c- 2a= 0 or c= 2a. It does not have a solution is c is not 2a."

--> Can you please explain how you got that from the rref? I see x+ (2/3)z, but where does the 2bx come from...?
 
  • #7
Row reduce the augmented matrix just as you would if were solving the equation- which is, after all the whole point of the problem.
 
  • #8
Oh ok, I understand. So, the augmented matrix had only numbers and b's on the right, where did the c and a variables come from?

This is the last thing I need to graps this. Thanks once again.
 
  • #9
can halls or anyone pick up on my last question?

Thank you.
 
  • #10
For some reason, I switched from <bx, by, bz> to <a, b, c> at the minute! I may have been doing the calculations on paper using a, b, c.
 
  • #11
OK, I see. Just for closure.

Since it is only if b_z is not 2b_x,

2
1
9

would be one such vector, right?

Thanks!
 

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their representations in vector spaces. It involves the use of algebraic operations and techniques to analyze and solve systems of linear equations and study the properties of vector spaces.

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