How is row elimination getting rid of this entry? This is a really elementary question, but I want to make sure I'm not missing something conceptual. In Strang's book Linear Algebra and Its Applications, on p. 321, he introduces tests for positive definiteness and semidefiniteness. As an example he shows that the matrix $$ \left [ \begin {matrix}  2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end {matrix} \right ]$$
is positive semi-definite. One criterion he uses is that no pivots are negative. To demonstrate this, he writes:

$$A = \left [ \begin {matrix}  2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end {matrix} \right ] \rightarrow \left [ \begin {matrix}  2 & 0 & 0 \\ 0 & \frac32 & -\frac32 \\ 0 & -\frac32 &  \frac32 \end {matrix} \right ] \rightarrow \left [ \begin {matrix}  2 & 0 & 0 \\ 0 & \frac32 & 0 \\ 0 & 0 & \mathbf{0}\end {matrix} \right ] \text { (missing pivot).}$$

How has he gotten from step 2 to step 3? Row elimination would not get rid of the $-\frac32$ in slot $2,3$ would it? I just want to make sure I correctly understand row elimination and how to uncover the pivots.
 A: Strang's book does have many typos and even has some outright errors, so one shouldn't be too hard on oneself if something doesn't quite make sense.  In fact I find Strang's book to be very confusing and difficult to understand even though I know a lot of linear algebra.  In my opinion the best way to study out of Strang is to have a clearer textbook handy to help decipher what he is saying.
That said, he gets the third matrix from the second by adding row 2 to row three to get all the zeros on the bottom row; then adding column 2 to column 3 to get all zeros in the 3rd column.  (This last step is unnecessary, but whatever).
That leaves you with 2 positive pivots and one zero pivot, which fits the pivot based definition of positive semi-definite.  I'm not a great fan of pivots, as I find them a little confusing.  I would translate the 3rd matrix as meaning that  the original matrix has one zero and two positive eigenvalues. One way of defining positive semi-definite is that all the eigenvalues are non-negative, at least one is zero, and at least one is > 0.  This is equivalent to any other definition of positive semi-definite, and I think it is the easiest of all of them to understand.
A: It seems likely that Strang was suggesting a congruency transformation rather than row equivalence. For $A$ symmetric, $A = L D L^T$ for $D$ diagonal, and we can look at the entries of $D$ to determine positive (semi-) definiteness.
