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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Please Help. Matrix with variables

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