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So I'm supposed to decide which $x \in \mathbb{R}$ gives a non-zero determinant:

\begin{vmatrix}1&x&x&2\\x&1&2&x\\x&2&1&x\\2&x&x&0\end{vmatrix}

It works well when when I re-write the matrix using Gaussian Elimination, only adding/subtracting rows to other rows. However, out of interest, I tried adding two rows to each other "simultaneously", because I suspected that it would bring me to the solution quicker. Basically, I tried subtracting row 3 with row 2, and row 2 with row 3 and replace them "at the same time". It's kind of difficult to explain, so I'll show you instead:

$$\begin{vmatrix}1&x&x&2\\x&1&2&x\\x&2&1&x\\2&x&x&0\end{vmatrix}=\begin{vmatrix}1&x&x&2\\0&-1&1&0\\0&1&-1&0\\2&x&x&0\end{vmatrix}=\begin{vmatrix}1&x&x&2\\0&0&0&0\\0&1&-1&0\\2&x&x&0\end{vmatrix}=0$$ Apparently this suggests that all values on $x$ produces a zero-valued determinant, which isn't true. I'm suspecting that there is something wrong in cross-adding/subtracting rows to other rows, but I can't quite figure out why. My wild guess is that there's some equivalence to multiplying a row by zero.

In order to pin-point this, I tried to compare it to a set of equations:

$$\begin{cases} x + y = 3 \\ 2x + y = 5\end{cases}$$

From this you obtain the matrix:

$$\begin{pmatrix}1&1&3\\2&1&5\end{pmatrix}\sim \begin{pmatrix}1&0&2\\0&1&1\end{pmatrix}$$ Which gives the solution $\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$

However, if I multiply row 1 with 2 and subtract it from row 2, and at the same time multiply row 2 with 1/2 and subtract that to row 1, I obtain: $$\begin{pmatrix}0&1/2&1/2\\0&-1&-1\end{pmatrix}\sim\begin{pmatrix}0&1&1\\0&0&0\end{pmatrix}$$

So basically, I've lost information about $x$, which makes sense. However, what bothers me is that I get confused when this is applied generally, such as in the case with the determinant. My questions are therefore:

  1. In what way does this type of "cross-addition/cross-subtraction" of matrices lose information? Maybe a vague question, but is there some way of seeing where/how the information is lost? Can it be compared to multiplying by 0 somehow?
  2. Is there any general rule for how you cannot perform operations on rows in matrices to avoid losing information? I know that you can't multiply a whole row by 0, but are there other things to look out for?

Basically, I'm just curious about this, since it hasn't been mention anywhere in my book or on any lecture. Maybe the answer is obvious and I'm just blind, but even if that's the case, I'd appreciate some guidelines!

Thanks a head!

EDIT: From experimenting a bit, I've started to realize that doing this cross-addition is similar to completely removing a row, which is similar, and even equivalent, to multiplying by 0. (That's why you always end up with at least one 0-row, or if you continue, a 0-matrix.) Maybe that was the connection I was looking for? It seems to make sense, but I may have missed something.

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  • $\begingroup$ You can't do simultaneous row additions because row reduction is a step by step process, and those steps cannot compose to give a simultenous row addition/subtraction in the way you are suggesting ($R3\to R3-R2$ and $R2\to R2-R3$). For a simple sanity check, show that if one is allowed to do simultaneous row additions, as well as the usual row operations, then every $2\times 2$ matrix can be reduced to the zero matrix. $\endgroup$
    – Dan Rust
    Oct 2, 2016 at 17:03
  • $\begingroup$ +Dan Rust Thanks for your response! I really appreciate it! I will try to show that. I understand everything you're saying, but I'm still getting confused whenever I try to think of what really happens when you do it simultaneously. Maybe the reduction you're talking about can help me see that better. $\endgroup$
    – Max
    Oct 2, 2016 at 17:15
  • $\begingroup$ Alright, I'm definitely getting more of it now than I did before. I edited the question to add my new conclusions, that I think may be correct. $\endgroup$
    – Max
    Oct 2, 2016 at 17:33
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    $\begingroup$ If you have performed $r_{2} \to r_{2} - r_{3}$ and $r_{3} \to r_{3} - r_{2}$ at the same time, then notice that the new rows $2$ and $3,$ say $r_{2}^{\prime}$ and $r_{3}^{\prime}$, are bound to satisfy $r_{2}^{\prime} = -r_{3}^{\prime}$, so the new matrix will always have linearly dependent rows ( hence zero determinant if square), whether or not the original matrix did. $\endgroup$
    – Geoff Robinson
    Oct 2, 2016 at 17:43

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