Understanding Solutions to Systems of Linear Equations

Click For Summary
The discussion revolves around solving a system of linear equations represented by a matrix. The equations derived from the matrix indicate that x must be 0, which leads to y also being 0, leaving z as the free variable. Participants explore different approaches to identify free variables, with some suggesting that either x or y can be treated as free variables depending on the context. The confusion arises in determining which variable to choose as free, highlighting the need for clarity in understanding the relationships between the variables. Ultimately, the key takeaway is recognizing that z is the only free variable in this specific system.
roam
Messages
1,265
Reaction score
12

Homework Statement



This is from a solved problem:

Here is a system:

\begin{bmatrix} {1 & 0&0 \\ -1&0&0\\3&-5&0 \end{bmatrix} \begin{bmatrix} {x \\ y\\z \end{bmatrix}= \begin{bmatrix} {0\\ 0\\0\end{bmatrix}

A general solution of this system is

x= \begin{bmatrix} {0 \\ 0\\t \end{bmatrix}= \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} t

So, how did they get this solution? I tried solving this and I just don't get it!

The Attempt at a Solution



The matrix corresponds to the set of equations:

x=0
-x=0
3x-5y=0

If I take "y" to be the free variable, I have x=5/3y therefore the solution to the system will be

\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {5/3 \\ 1\\0 \end{bmatrix} y + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z

But if I take "x" to be the free variable I get y=3/5x, so I get the following solution:

\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {1 \\ 3/5\\0 \end{bmatrix} x + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z

How do I know which one (x or y) should be taken as the free variable?

Could anyone please explain to me how to get the right answer? :frown: I just can't see how to deal with this type of problems.
 
Physics news on Phys.org
roam said:

Homework Statement



This is from a solved problem:



So, how did they get this solution? I tried solving this and I just don't get it!

The Attempt at a Solution



The matrix corresponds to the set of equations:

x=0
-x=0
3x-5y=0
From these equations, x is obviously 0. With x = 0, the 3rd equation implies that y = 0. This means that z is the free variable.
roam said:
If I take "y" to be the free variable, I have x=5/3y therefore the solution to the system will be

\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {5/3 \\ 1\\0 \end{bmatrix} y + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z

But if I take "x" to be the free variable I get y=3/5x, so I get the following solution:

\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {1 \\ 3/5\\0 \end{bmatrix} x + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z

How do I know which one (x or y) should be taken as the free variable?

Could anyone please explain to me how to get the right answer? :frown: I just can't see how to deal with this type of problems.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
Replies
19
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
10
Views
2K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K