LDL^T factorization of a tridiagonal matrix giving wrong result

[stunner5000pt]

Find the $LDL^T$ factorization of this matrix

$$\left(\begin{array}{ccc}{2&-1&0\\-1&2&-1\\0&-1&2\end{array}\right)$$

now i can find the L matrix by gaussian elimination
that yields
$$L = \left(\begin{array}{ccc}{1&0&0\\\frac{-2}{3}&1&0\\0&\frac{-1}{2}&1\end{array}\right)$$
$$D = \left(\begin{array}{ccc}{\frac{1}{4}&0&0\\0&\frac{1}{3}&0\\0&0&\frac{1}{2}\end{array}\right)$$

i am pretty sure about the ansswer since i checked my working many times.
However this is not the answer at the back of the book! In fact i am not even close!
What am i doing wrong?? Can anyone please help me iwth this?
Thank you for your help!

 

[HallsofIvy]

It's hard to tell what you did wrong when you did wrong when you don't tell us what you did! I did a quick "column reduction" to get L and didn't get any like you got.

 

[stunner5000pt]

well i got those answers by Gaussian Elimination
this is what i did

$$\left(\begin{array}{ccc}{2&-1&0\\-1&2&-1\\0&-1&2\end{array}\right)$$

R3 + 2R2
$$\left(\begin{array}{ccc}{2&-1&0\\-2&3&0\\0&-1&2\end{array}\right)$$

R2+3R1
$$\left(\begin{array}{ccc}{4&0&0\\-2&3&0\\0&-1&2\end{array}\right)$$

and my textbook says that that the D matrix is formed by dividing the square terms of the lower matrix formed and multiply that by the elementary matrix yielding
$$D = \left(\begin{array}{ccc}{\frac{1}{4}&0&0\\0&\frac{ 1}{3}&0\\0&0&\frac{1}{2}\end{array}\right)$$

 

[stunner5000pt]

can anyone tell me what i have done wrong? my answer is not even close to the tedxxt book's answer. However all my steps with the row reductions are correct, as you can see.

I was told that i was not supposed to use row reduction to get the lower matrix? SO what do i do then?