Please Help. Matrix with variables

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This is a pre-calculus problem. I don't know if I got the answer right, and I don't quite understand what I'm doing. I read the section several times and understand regular augmented matrices but this problem is confusing me a bit. There is no mention in the text about matrices with variables in the matrices. There is no answer in the back of the book.

Find this matrix in rref form and for what values of x is this valid.

matrix[1 x -1] My work so far:[x 1 -1] Switched order of rows
[x 1 -1] [1 x -1]
[-1 1 x] [-1 1 x] (continues below and to the left)



[0 1-x^2 x-1] added multiples of middle row to other rows
[1 x -1]
[0 x+1 x-1]

Added multiple of top row to middle row, x can't equal 1 or -1 (multiple is x/[(x +1)(x-1)])

[0 1-x^2 x-1]
[1 0 (x/(x+1)) -1]
[0 x+1 x-1]


[0 -(x^2-1) x-1]
[1 0 -1/(x+1)] simplify
[0 x+1 x-1]

[1 0 -1/(x+1)] Simplify to rref. Again, x can’t be 1 or -1
[0 1 -1/(x+1)]
[0 1 (x-1)/(x+1)]


the three systems at this point:

a = -1/(x+1)
b = -1/(x+1)
b = (x-1)/(x+1)

Setting the last to = to each other i get

(1/(x-1)) times (-1-(x-1)) = 0

and x=0, x doesn't equal 1 again


a= -1 and b=-1 plugging in

so i get

[1 0 -1]
[0 1 -1]
[-1 1 0]

and through addition of the first row to the third row

[1 0 -1]
[0 1 -1]
[0 1 -1] for the rref equivalent matrix


I FIND IT WEIRD THAT THE ONLY NUMBER X COULD BE IS 0. AM I WRONG??

Why did the creator of this problem put "x" in the matrix. If you have three variables and three equations, why not make a 3 by 4 augmented matrix?

If I write this out

a + bx = -1
ax + b = -1
-a + b = x

IT DOESN'T SEEM LIKE A LINEAR EQUATION? WHY IS IT BEING SOLVED WITH "LINEAR ALGEBRA." Is X only allowed to be a single number (an unknown coefficient)?

Thank you for your help
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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This is a pre-calculus problem. I don't know if I got the answer right, and I don't quite understand what I'm doing. I read the section several times and understand regular augmented matrices but this problem is confusing me a bit. There is no mention in the text about matrices with variables in the matrices. There is no answer in the back of the book.

Find this matrix in rref form and for what values of x is this valid.

matrix[1 x -1] My work so far:[x 1 -1] Switched order of rows
[x 1 -1] [1 x -1]
[-1 1 x] [-1 1 x] (continues below and to the left)


Your initial matrix is

[1 x -1]
[x 1 -1]
[-1 1 x]

?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
This is a pre-calculus problem. I don't know if I got the answer right, and I don't quite understand what I'm doing. I read the section several times and understand regular augmented matrices but this problem is confusing me a bit. There is no mention in the text about matrices with variables in the matrices. There is no answer in the back of the book.

Find this matrix in rref form and for what values of x is this valid.

matrix[1 x -1] My work so far:[x 1 -1] Switched order of rows
[x 1 -1] [1 x -1]
[-1 1 x] [-1 1 x] (continues below and to the left)
So the original matrix is
[tex]\left[\begin{array}{ccc} 1 & x & -1 \\ x & 1 & -1 \\ -1 & 1 & x\end{array}\right]


[0 1-x^2 x-1] added multiples of middle row to other rows
[1 x -1]
[0 x+1 x-1]

Added multiple of top row to middle row, x can't equal 1 or -1 (multiple is x/[(x +1)(x-1)])

[0 1-x^2 x-1]
[1 0 (x/(x+1)) -1]
[0 x+1 x-1]


[0 -(x^2-1) x-1]
[1 0 -1/(x+1)] simplify
[0 x+1 x-1]

[1 0 -1/(x+1)] Simplify to rref. Again, x can’t be 1 or -1
[0 1 -1/(x+1)]
[0 1 (x-1)/(x+1)]


the three systems at this point:

a = -1/(x+1)
b = -1/(x+1)
b = (x-1)/(x+1)

Setting the last to = to each other i get

(1/(x-1)) times (-1-(x-1)) = 0
Why would you want them to "equal each other"? There is nothing like that in the problem. Clearly the rref form for this is
[tex]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & \frac{x-1}{x+1}\end{array}\right][/tex]
or simply to
[tex]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{array}\right][/tex]

as long as x is NOT either 1 or -1.

and x=0, x doesn't equal 1 again


a= -1 and b=-1 plugging in

so i get

[1 0 -1]
[0 1 -1]
[-1 1 0]

and through addition of the first row to the third row

[1 0 -1]
[0 1 -1]
[0 1 -1] for the rref equivalent matrix


I FIND IT WEIRD THAT THE ONLY NUMBER X COULD BE IS 0. AM I WRONG??
x is 0 in order that what happens? The problem, as you stated it, was simply to row reduce the matrix. That can be done for any x except 1 or -1.

Why did the creator of this problem put "x" in the matrix. If you have three variables and three equations, why not make a 3 by 4 augmented matrix?

If I write this out

a + bx = -1
ax + b = -1
-a + b = x

IT DOESN'T SEEM LIKE A LINEAR EQUATION? WHY IS IT BEING SOLVED WITH "LINEAR ALGEBRA." Is X only allowed to be a single number (an unknown coefficient)?

Thank you for your help
Nothing was said about equations. A matrix in general does not have to be the "augmented" matrix of any set of equations.
 
  • #4
5
0
Thank you that was very helpful.
 

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