Below is a sketch of how one often bounds the number of operations needed, by comparing the total number of distinct matrices (namely $2^{n^2}$) against how many matrices you can reach with a given number of operations. The key idea is that each operation set—row‐additions mod 2 plus “flip‐two‐bits”—generates a certain “branching factor,” and one obtains a lower bound by insisting that the reachable set of matrices must be large enough to include all $2^{n^2}$ possibilities.

1. The setting

• We have an $n \times n$ binary matrix $M$ (entries in $\{0,1\}$).
• We want to get from the all‐zeros matrix $0$ to an arbitrary matrix $M$ using two types of operations:
  1. Row‐add mod 2: Pick two rows $r_i$ and $r_j$ (possibly $i \neq j$), and replace one of them by the sum mod 2: $$r_i \;\leftarrow\; r_i \;+\; r_j \pmod 2.$$ Equivalently, for each column $k$, $$(r_i)_k \;\leftarrow\; (r_i)_k \oplus (r_j)_k.$$
  2. Flip two bits: Pick any two entries of the matrix and flip them (0 to 1 or 1 to 0).

We want a handle on the minimum number of operations needed to transform the all‐zeros matrix into any of the $2^{n^2}$ possible binary matrices. Or, equivalently, we can ask: if we allow ourselves $k$ operations, how many distinct matrices can we reach?

Because there are $2^{n^2}$ total matrices, to be able to reach all of them, we need to ensure that the number of reachable matrices (under at most $k$ operations) is at least $2^{n^2}$. This is a common counting (“sphere‐covering” or “branch‐factor”) argument.

2. Counting reachable matrices versus total matrices

1. Total matrices: There are clearly

   $$2^{n^2}$$

   different $n\times n$ binary matrices.

2. Bounding the number of reachable matrices with $k$ operations:

   • At each step, you have at most two kinds of moves:

     1. Row‐add mod 2:
        You must choose an ordered pair $(i,j)$ of distinct rows (or possibly allow $i=j$, but that would be trivial since adding a row to itself is no change). The number of distinct ways to choose $(i,j)$ can be on the order of $n^2$ if you consider all pairs, but more precisely it is at most $n(n-1)$ if $i\neq j$.

        However, importantly, if you start from the all‐zeros matrix, note that applying row‐add does nothing initially (since adding the zero row changes nothing). So to get non‐trivial states, you typically need to do some flips first. Nonetheless, from a purely combinatorial upper‐bound perspective, each state can branch into at most $O(n^2)$ new states by row‐addition (some may be duplicates, but we just need an upper bound).

     2. Flip two bits:
        You must choose 2 positions (out of $n^2$) to flip. There are $\binom{n^2}{2} = O(n^4)$ ways to pick which bits to flip. Each such choice might lead to a different next matrix.

   • In the worst case (for an upper bound), you can say that each operation has at most

     $$C \;=\; \binom{n^2}{2} \;+\; n(n-1)$$

     possible ways to move to a new matrix state, for some constant factor in front. Realistically, $\binom{n^2}{2}$ can dominate $n(n-1)$ for large $n$, so $C$ can be taken to be on the order of $n^4$.

   • Therefore, after $k$ operations, a very rough upper bound on the number of distinct matrices you can reach is

     $$\text{(\# reachable states)} \;\le\; 1 \;+\; C \;+\; C^2 \;+\;\cdots+\; C^k \;\le\; \frac{C^{\,k+1}-1}{C-1} \;\approx\; C^{\,k}.$$

     (We started from 1 state: the all‐zeros matrix. Then each step fans out by at most $C$ possibilities.)

3. Comparing to $2^{n^2}$: We want enough operations $k$ so that the total number of distinct matrices reachable is at least $2^{n^2}$. From

   $$C^k \;\ge\; 2^{n^2},$$

   we get

   $$k \;\ge\; \frac{n^2 \ln 2}{\ln C}.$$

   Since $C \approx n^4$ for large $n$, this implies roughly

   $$k \;\ge\; \frac{n^2 \,\ln 2}{\ln(n^4)} \;=\; \frac{n^2 \,\ln 2}{4\,\ln(n)}.$$

   That gives you a coarse asymptotic lower bound.

   Of course, one can refine the argument (for instance, flipping bits in a controlled manner and then using row‐adds), but this counting approach already shows you cannot hope to reach all of the $2^{n^2}$ matrices unless $k$ grows on the order of at least $\frac{n^2}{\ln(n)}$. In fact, many more careful arguments end up concluding you likely need a linear (in $n^2$) number of bit‐flips (since row‐add alone does not create new 1s from the zero matrix).

3. Takeaways

1. Row‐add mod 2 is a linear operation in GF(2). From the zero matrix, it does nothing unless you have already introduced some 1s via flips.
2. Flipping two bits can introduce new 1s, but each flip “costs” you an operation.
3. The branching argument says that with $k$ operations you can reach at most on the order of $C^k$ distinct matrices (where $C$ is polynomial in $n$).
4. Since there are $2^{n^2}$ matrices total, you need $C^k \ge 2^{n^2}$. Solving for $k$ yields a lower bound of order $\frac{n^2}{\log(n)}$ in the simplest counting estimate (or stronger, depending on more refined arguments).

Hence, even though row‐add mod 2 is powerful in certain senses (it easily “moves” bits around rows if they already exist), introducing those bits requires enough flips that your number of operations must grow at least on the order of $n^2$ for large matrices.

Below is an outline of the standard counting argument (sometimes called a “branching” or “sphere‐covering” argument) for bounding how many operations are needed when the allowed operations are:

1. Add two rows mod 2 (pick rows $i \neq j$ and do $r_i \leftarrow r_i + r_j \pmod{2}$).
2. Flip a single bit (pick one entry in the matrix and flip 0 to 1 or 1 to 0).

We want to transform the all‐zeros $n \times n$ matrix into any possible binary matrix. Since there are $2^{n^2}$ total binary matrices, we ask: How large must $k$ be so that, in at most $k$ operations, we can reach all $2^{n^2}$ matrices? Equivalently, we get a lower bound on $k$ by requiring that the number of distinct matrices reachable in $\le k$ steps is at least $2^{n^2}$.

1. The total number of matrices

Clearly, there are

possible $n \times n$ binary matrices (since each of the $n^2$ entries can be $0$ or $1$ independently).

2. Upper‐bounding the number of reachable matrices

1. Start: We begin at the all‐zeros matrix.

2. Operations:

   • Row‐add mod 2: We pick two distinct rows $i, j$ (with $i \neq j$) and perform

     $$r_i \;\leftarrow\; r_i \;+\; r_j \pmod{2}.$$

     There are at most $n(n-1)$ ways to choose the ordered pair $(i,j)$.

   • Flip a single bit: We pick exactly 1 entry out of $n^2$ and flip it (0 to 1 or 1 to 0). There are $n^2$ ways to do this in any given matrix state.

3. Branching factor:
   If we denote by $C$ an upper bound on the number of distinct next‐matrices you can get from a single matrix in one operation, then

   $$C \;\le\; n(n-1) \;+\; n^2 \;\approx\; 2\,n^2$$

   for large $n$. (Some of those moves might coincide or lead to the same resulting matrix, but for an upper bound we can just count them all.)

4. Number of reachable matrices in $\le k$ steps:
   A rough upper bound on the number of distinct matrices reachable by at most $k$ operations is

   $$1 + C + C^2 + \cdots + C^k \;\le\; \frac{C^{k+1} - 1}{C - 1} \;\approx\; C^k \quad (\text{for large }k).$$

   We started from exactly 1 matrix (the zero matrix). After each operation, we can branch to $\le C$ new states, etc.

Hence

3. Matching or exceeding $2^{n^2}$

To be able to reach all $2^{n^2}$ matrices (in particular, to have at least that many reachable states), we need

Taking logarithms gives

For large $n$, $\ln(2\,n^2) \approx \ln(n^2) = 2\ln(n)$. Thus asymptotically,

So this simple counting argument shows that at least on the order of $\tfrac{n^2}{\ln n}$ operations are needed if one hopes to reach all $2^{n^2}$ matrices starting from the zero matrix.

Key idea: Even though “row‐add mod 2” can rearrange or combine already‐existing 1s in interesting ways, it does not create new 1s out of nowhere. To introduce 1s, you must use “flip a single bit.” Because each operation only flips one bit, you cannot create too many distinct states too quickly. Hence, counting how many total states you can branch into at each step yields a lower bound on the number of operations required to cover all matrices.

4. Takeaways

1. Total state space is $2^{n^2}$.
2. Branching: In each operation, you either (a) choose one of $n(n-1)$ row‐adds, or (b) choose one of $n^2$ single‐bit‐flips—so $\le 2n^2$ moves per step in a crude upper bound.
3. Reachable states within $k$ steps is then at most $(2n^2)^k$.
4. Compare $(2n^2)^k$ to $2^{n^2}$ and solve for $k$. You get $k \ge \Omega\!\bigl(n^2/\ln(n)\bigr)$.

In short, you need at least on the order of $\frac{n^2}{\ln(n)}$ operations to ensure you can transform the zero matrix into any $n \times n$ binary matrix using “row‐add mod 2” and “flip exactly one bit.”

Below is a more precise version of the standard counting (branching) argument. We will derive an explicit formula for the lower bound on the number of operations needed to reach all $2^{n^2}$ binary $n\times n$ matrices from the all‐zero matrix, when the allowed operations are:

1. Add two rows mod 2: choose $i\neq j$ and replace row $i$ by $r_i \leftarrow r_i + r_j \pmod{2}.$

2. Flip a single bit: choose one cell in the matrix and toggle it (0 → 1 or 1 → 0).

1. Counting possible moves from a single matrix

From any given matrix $A$, you have at most

1. Row‐add moves:
   There are $n(n-1)$ possible ordered pairs $(i,j)$ with $i\neq j$, so at most $n(n-1)$ ways to “add row $j$” into row $i$ mod 2.

2. Single‐bit‐flip moves:
   There are $n^2$ entries, so $n^2$ ways to choose exactly one bit to flip.

Hence from a given matrix $A$, there are at most

distinct “next” matrices you can obtain in one operation. (In practice, some moves might coincide, but this upper‐count still suffices for a worst‐case branching bound.)

2. Counting how many matrices you can reach in $k$ steps

1. You start with exactly 1 matrix (the all‐zeros matrix).

2. In each step, from each matrix in your current “reachable set,” you can move to at most $M$ new matrices.

3. Therefore, in at most $k$ steps, the maximum size of your reachable set is bounded above by

For large $k$, the dominant term is $M^k$. To keep it clean, one can say a simpler upper bound is

3. Reaching all $2^{n^2}$ matrices

We have a total of $2^{n^2}$ distinct $n\times n$ binary matrices. If we want to be able to reach all of them (or at least have the capacity to reach that many distinct states), then we need

Recall $M = 2n^2 - n$. Take the natural log of both sides:

Hence

This is an exact inequality for a simple geometric‐branching upper bound.

3.1 Asymptotic approximation

For large $n$, $\ln(2n^2 - n)$ is approximately

Thus

Plugging this back in gives

So for large $n$,

In more explicit constant terms, for large $n$:

since $\ln(2)\approx 0.693$. The precise constant is not crucial for big‐$n$ lower bounds (we just care that it grows like $n^2/\ln(n)$).

4. Take‐home statement

1. Exact form of the bound from this simple “branching” argument: $$k \;\;\ge\;\; \frac{n^2\,\ln(2)}{\ln\bigl(2n^2 - n\bigr)}.$$
2. Asymptotic form for large $n$: $$k \;=\; \Omega\!\Bigl(\tfrac{n^2}{\ln(n)}\Bigr).$$

Either way, this shows that at least on the order of $n^2 / \ln n$ operations are needed if you want to be able to reach every possible $n\times n$ binary matrix (all $2^{n^2}$ of them), starting from the zero matrix, using only:

• “Add row $j$ to row $i$$ (mod 2)”
• “Flip exactly one bit at a time.”

Below is some insight into why there must be a specific matrix requiring $\Omega\!\bigl(\tfrac{n^2}{\ln n}\bigr)$ operations—and how one might construct or exhibit such a matrix, at least with high probability. The short summary is:

1. A pigeonhole (counting) argument ensures some matrix is unreachable in fewer than that many steps.
2. A probabilistic method can make this argument more concrete by showing that a randomly chosen matrix will, with high probability, require that many steps.

In most treatments, the existence of a “hard-to-reach” matrix is non-constructive: we know at least one must exist, but we don’t necessarily know which one without further machinery. However, there are ways (often using coding theory or combinatorial designs) to explicitly build such matrices as well.

1. Pigeonhole / counting argument (existence)

From the counting argument, we see that with $k$ operations allowed, you can reach at most

distinct matrices (starting from the zero matrix).

But there are $2^{n^2}$ total binary $n\times n$ matrices. If

then not all matrices can be reached with $\le k$ operations. In particular, this means there exists at least one matrix $M_{\text{hard}}$ such that transforming the zero matrix into $M_{\text{hard}}$ requires more than $k$ operations.

Solving $(2n^2 - n)^k \approx 2^{n^2}$ shows $k \gtrsim \tfrac{n^2}{\ln(n)}$. Hence, by pigeonhole principle alone, some matrix requires on the order of $\tfrac{n^2}{\ln n}$ operations.

Key point: This argument does not immediately tell you which matrix is the hardest—it just guarantees that at least one matrix is that hard.

2. Making a “hard” matrix via the probabilistic method

Although the basic pigeonhole argument shows existence, one can often make that argument more concrete by using a random construction:

1. Pick a random matrix $M$ of size $n\times n$ uniformly from $\{0,1\}^{n^2}$.

2. Show that the probability that $M$ can be formed in $\le k$ operations is very small when $k < c\,\tfrac{n^2}{\ln n}$.

Hence most random matrices (i.e., with high probability) are “hard.” This approach is still not giving you a single, explicit matrix by formula, but it gives a probabilistic guarantee that if you pick a matrix at random, it very likely needs many operations to produce.

2.1 Sketch of that argument

• There are at most $(2n^2 - n)^k$ outcomes reachable in $\le k$ operations.

• The total number of possible matrices is $2^{n^2}$.

• If

  $$(2n^2 - n)^k \;\ll\; 2^{n^2},$$

  then the fraction of matrices reachable in $\le k$ operations is small. Equivalently, a random matrix has a small chance (much less than 1) of being in the set of “reachable in $\le k$ steps.”

Thus, most matrices are not reachable in $\le k$ steps, meaning most require more than $k$ operations.

3. Obtaining an explicit hard matrix

Sometimes one wants an explicit (deterministically described) matrix that needs $\Omega\!\bigl(\tfrac{n^2}{\ln n}\bigr)$ operations. Typically, this is done by more refined combinatorial tools or coding‐theory–style arguments:

1. Coding ideas:

   • Imagine encoding each possible “short sequence of operations” as a “codeword.” If you can show that code has limited distance properties or certain constraints, you can build or find a matrix that is far from all codewords. In effect, you show that every short sequence fails to produce your special matrix.

2. Expander / design arguments:

   • Sometimes, you can use expanders or design‐theoretic arguments to force that any short sequence of bit‐flips plus row‐adds can’t cover certain large sets of bits. You then construct a matrix that has 1s exactly in a “difficult arrangement” that short sequences can’t realize.

These methods get quite technical, and the typical approach in a lecture or a paper is simply to assert the simpler existence or probabilistic approach—especially if the main point is just that we need at least $\Omega(n^2 / \ln n)$ operations in the worst case.

4. Summary

• From the counting argument, we know there must be a matrix requiring $\Omega(n^2 / \ln n)$ row‐adds/single‐bit‐flips to obtain from the zero matrix.

• By the probabilistic method, we see that most matrices (chosen at random) in fact require that many operations (with high probability).

• If you truly want a concrete, explicit example of a matrix needing that many steps, you typically use more advanced combinatorial or coding‐theory arguments. Such proofs show how to construct a matrix with particular properties that rule out any short sequence of operations.

In day‐to‐day mathematics, it’s common to be satisfied knowing at least one matrix (indeed, almost all matrices) is that hard, without necessarily pinpointing exactly which one.

Below is an outline of how one can give a fully explicit (deterministically described) family of $n\times n$ matrices over $\{0,1\}$ that each require on the order of $\tfrac{n^2}{\log n}$ row‐additions and single‐bit‐flips to obtain from the zero matrix. The high‐level idea is to exploit a known type of combinatorial design or expander‐graph–based construction whose “global” 1‐pattern is impossible to build from the zero matrix with few flips and XOR‐ing of rows.

Most published proofs of such lower bounds either

1. Show non‐constructively (by counting or by a probabilistic argument) that almost every matrix is this hard, or
2. Reference more advanced combinatorial machinery (codes, expanders, etc.) to pin down a concrete example.

Below, we sketch how an expander‐based adjacency matrix (or certain high‐distance error‐correcting codes) can serve as an explicit worst‐case example.

1. Why it is tricky

Most “simple” matrices you might guess (like having 1s on a diagonal, or 1s in a small pattern) can be built fairly quickly via row‐XOR operations plus a small number of flips. Proving that a particular matrix $M$ cannot be formed in fewer than $\alpha\,\frac{n^2}{\log n}$ steps usually requires showing that any short sequence of bit‐flips and row‐adds leaves behind some global “signature” that fails to match $M$.

That in turn often relies on:

1. Distance or expansion properties of $M$.
2. A combinatorial lemma bounding how quickly a small number of flips plus XORs can “spread” 1s throughout the matrix.

Hence, while almost all matrices are that hard, exhibiting a specific matrix with a clear pattern typically calls on expander graphs or error‐correcting codes.

2. The expander‐graph (bipartite adjacency) construction

A fairly standard explicit construction uses a bipartite expander $G = (U,V,E)$ with $|U|=n$, $|V|=n$. One arranges $G$ so that:

• Each node $u \in U$ has (say) $d$ neighbors in $V$.
• Every small subset $S \subseteq U$ has at least $\rho \cdot |S|$ distinct neighbors in $V$, for some expansion factor $\rho\in (0, d)$.
• There are known, fully explicit constructions of such expanders—e.g., via algebraic (Ramanujan) graphs, or certain combinatorial/multigraph constructions running in polynomial time.

We then form an $n\times n$ adjacency matrix $M$ by setting

2.1. Why is $M$ hard to build from the zero matrix?

• A single flip of a bit creates exactly one new “1” in some position $(i,j)$.
• A row‐add (XOR of row $r_j$ into row $r_i$) toggles exactly those columns in row $i$ where row $j$ has a 1.

Roughly speaking, to “populate” row $i$ with the correct adjacency pattern (i.e.\ the correct set of 1’s) you must already have introduced those 1’s somewhere else to copy them by XOR. But an expander can be chosen so that each row has many 1’s, and different rows have 1’s in somewhat disjoint sets of columns. Thus, each row effectively needs many new flips if you do not already have them “available” in some other row. One can formalize this so that any sequence of fewer than $c\,\tfrac{n^2}{\log n}$ flips plus row‐XORs fails to reach the global 1‐pattern of the expander adjacency matrix.

Such a proof typically rests on a lemma of the flavor:

Expander Lemma (informal).
If each row of $M$ has at least $d$ ones and each small group of rows collectively covers a large number of columns, then any attempt to generate these 1s from only $k$ flips + row‐XORs will fall short unless $k$ is large (roughly $\Omega(\tfrac{n^2}{\log n})$).

One can make this rigorous by bounding how fast new 1’s can “spread” into multiple columns, given that each row‐XOR toggles entire columns at once. With an expander’s guarantee that subsets of rows have big neighborhoods, you force any small set of flips to leave many rows incorrectly zeroed out in too many columns.

2.2. Bottom line

Because good expanders can be built explicitly (there are well‐understood polynomial‐time constructions), the resulting adjacency matrices are a deterministic, “closed‐form” family of matrices that meet the lower bound.

3. An alternate code‐based construction

Another common strategy uses error‐correcting codes of large minimum distance. For instance:

1. Let the matrix dimension be $n \times n$. Think of each row as a length‐$n$ binary vector in $\mathbf{F}_2^n$.
2. We choose these row vectors so that no small combination of them is close (in Hamming distance) to some target row—unless we do many bit‐flips. In short, flipping only a few bits plus taking XORs of existing rows cannot produce that row.
3. By carefully assembling all $n$ rows to have strong pairwise and subset‐wise distance properties, we engineer a single matrix $M$ that is “globally difficult” to obtain from an all‐zero start.

Explicit linear codes (e.g.\ BCH codes, Reed–Muller codes, or certain algebraic‐geometry codes) provide large distance and also allow explicit, deterministic generation of each row in time polynomial in $n$. One then proves that any short sequence of flips + row‐XORs fails to reconstruct the full matrix $M$.

The details can get somewhat intricate—one must show that from the zero matrix you cannot “assemble” all those carefully spaced codewords with fewer than $\Omega(\tfrac{n^2}{\log n})$ flips.

4. Concretely stating such a matrix

In practice, papers and textbooks often say something like:

“Let $M_n$ be the adjacency matrix of a suitably chosen $(n,d,\rho)$ bipartite expander (constructed via the known Ramanujan‐graph method of Lubotzky–Phillips–Sarnak or a standard Margulis–Gabber–Galil expander). Then $M_n$ is an $n\times n$ $\{0,1\}$‐matrix which requires at least $c\,\tfrac{n^2}{\log n}$ row‐additions and bit‐flips to obtain from the zero matrix, for some universal constant $c>0$.”

That statement is completely explicit: there is a well‐defined polynomial‐time algorithm (based on algebra over finite rings, or on “lifting” smaller expanders) to construct each $M_n$. One can write down a formula for the edges of the expander graph in closed form.

Likewise, using an explicit high‐distance code, you can say:

“Let $M_n$ be the matrix whose $i$th row is the codeword of a BCH code (or Reed–Muller code) of length $n$ corresponding to message index $i$. Then $M_n$ has the property that …”

and you proceed to prove the same kind of lower bound.

Because these expanders/codes are classical, the matrices $M_n$ are 100% constructive.

Final summary

1. Counting arguments show that at least one matrix needs $\Omega(\tfrac{n^2}{\log n})$ operations—non‐constructive.

2. Probabilistic arguments show that most random matrices need that many operations—still not an explicit recipe.

3. Explicit expanders or codes yield a concrete infinite family $\{\,M_n\}$ of $n\times n$ matrices with the required property. Each $M_n$ can be written down by a deterministic process (an algebraic formula or a simple polynomial‐time algorithm).

Thus, if you truly need to exhibit one specific “worst‐case” matrix for each $n$, the usual recommendation is:

Take the adjacency matrix of a known bipartite expander (Ramanujan or similar).

Proving rigorously that it indeed needs $\Omega(n^2/\log n)$ operations is then a (somewhat intricate) but standard argument in the theory of expanders versus “small XOR+flip” generating sets. This is how one obtains an explicit example.