1. Interpret the question

We need the number of $3\times 3$ matrices with:

• exactly four entries equal to $1$,
• all remaining five entries equal to $0$,
• and the matrix must be non-singular (i.e. determinant $\neq 0$).

So we must count all invertible $0$-$1$ matrices of size $3\times 3$ having exactly four $1$'s.

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2. Necessary condition for non-singularity

If any row or any column is entirely zero, then the determinant is $0$.

Since there are exactly four $1$'s distributed among $3$ rows, for a non-singular matrix each row must contain at least one $1$. Similarly, each column must contain at least one $1$.

Now $4$ ones are to be distributed among $3$ rows with each row having at least one $1$. Thus the row-sum pattern must be:

$$(2,1,1)$$

up to permutation.

Similarly, the column-sum pattern must also be:

$$(2,1,1)$$

up to permutation.

So every invertible matrix of this type must have exactly one row containing two $1$'s, and exactly one column containing two $1$'s.

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3. Count such matrices systematically

Choose:

• the row which has two $1$'s: $3$ choices,
• the column which has two $1$'s: $3$ choices.

Now fix such a row and such a column. We count the number of matrices with row sums $(2,1,1)$ and column sums $(2,1,1)$.

Let the special row be $R_i$ and the special column be $C_j$.

Because column $C_j$ must contain two $1$'s, and the other two columns contain one $1$ each, and row $R_i$ must contain two $1$'s, there are exactly two possible patterns:

Case 1: entry $(i,j)=1$

Then row $R_i$ needs one more $1$ in one of the other two columns. Also column $C_j$ needs one more $1$ in one of the other two rows. The remaining two rows and two columns then get forced uniquely. This gives 2 possibilities.

Case 2: entry $(i,j)=0$

Then row $R_i$ must have $1$'s in both the other columns, and column $C_j$ must have $1$'s in both the other rows. Again the matrix is uniquely determined. This gives 1 possibility.

Let us verify by taking a concrete example: suppose row 1 and column 1 are the ones with sum $2$. Then the possible matrices are

$$\begin{pmatrix} 1&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix}, \quad \begin{pmatrix} 1&0&1\\ 1&0&0\\ 0&1&0 \end{pmatrix}, \quad \begin{pmatrix} 0&1&1\\ 1&0&0\\ 1&0&0 \end{pmatrix}$$

But the third matrix is singular because rows 2 and 3 are equal. So we must check non-singularity carefully.

Thus the above raw count includes singular cases.

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4. Better approach: count non-singular ones directly

A $3\times 3$ matrix with exactly four $1$'s is singular if:

• some row/column is zero, or
• two rows are equal, or
• rows are linearly dependent.

Since row sums must be $(2,1,1)$, the two rows with a single $1$ must be in different columns, otherwise they become equal and determinant becomes $0$. Also, the row with two $1$'s must contain those two columns, and the remaining column must contain the lone $1$ of the third row.

So a non-singular matrix must have row patterns exactly of the form:

$$e_a+e_b,\quad e_a,\quad e_c$$

where $a,b,c$ are distinct and $\{a,b,c\}=\{1,2,3\}$. In other words:

• one row has ones in two columns,
• the other two rows have a single $1$ each,
• and those single $1$'s are in two different columns,
• one of those columns is shared with the double-$1$ row, and the third column is occupied by the remaining single-$1$ row.

A clean way to count:

• Choose the row having two $1$'s: $3$ choices.
• Choose the column where the row having two $1$'s has a $0$: $3$ choices. (equivalently, choose which two columns contain its $1$'s.)
• For the remaining two rows, place the two single $1$'s in the two columns not both same; to keep all columns nonzero and rows distinct, their positions are forced up to permutation between the two rows: $2$ choices.

Hence total number is

$$3\times 3\times 2 = 18.$$

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5. Check with examples

For example, if row 1 has two $1$'s and its zero is in column 3, then row 1 is

$$(1,1,0).$$

To make the matrix invertible, rows 2 and 3 must be

$$(1,0,0),\ (0,0,1)$$

or

$$(0,1,0),\ (0,0,1),$$

and these can be assigned to rows 2 and 3 in $2$ ways in each pattern? Let's test carefully.

Actually, for row 1 = $(1,1,0)$, the remaining two rows must be one of the following unordered pairs:

• $(1,0,0)$ and $(0,0,1)$,
• $(0,1,0)$ and $(0,0,1)$.

Each unordered pair can be arranged in rows 2 and 3 in $2$ ways, giving $4$ possibilities for this row pattern.

So my previous count of $18$ may be incomplete. Let us recount properly.

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6. Most reliable counting via row types

Choose the row with two $1$'s.

Step 6.1: Choose the row having two $1$'s

This can be done in

$$3$$

ways.

Step 6.2: Choose the pattern of that row

A row with exactly two $1$'s can be one of:

$$(1,1,0),\ (1,0,1),\ (0,1,1)$$

So there are

$$3$$

choices.

Step 6.3: Choose the other two rows

Suppose the chosen two-$1$ row is $(1,1,0)$. Then to make all columns nonzero and avoid dependence, the remaining two rows must be two distinct standard basis rows chosen from

$$(1,0,0),\ (0,1,0),\ (0,0,1)$$

such that together with $(1,1,0)$ they form an invertible matrix.

The valid pairs are:

• $(1,0,0)$ and $(0,0,1)$,
• $(0,1,0)$ and $(0,0,1)$.

Pair $(1,0,0)$ and $(0,1,0)$ is invalid because then column 3 is zero.

So there are $2$ valid unordered pairs, and each can be arranged in the two remaining row positions in

$$2!$$

ways.

Hence for a fixed two-$1$ row pattern and fixed row position, number of matrices is

$$2\times 2 = 4.$$

Therefore total number is

$$3\times 3\times 4 = 36.$$

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7. Sanity check with one example

Take row 1 = $(1,1,0)$. Then the four valid completions are

$$\begin{pmatrix} 1&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix}, \begin{pmatrix} 1&1&0\\ 0&0&1\\ 1&0&0 \end{pmatrix}, \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \begin{pmatrix} 1&1&0\\ 0&0&1\\ 0&1&0 \end{pmatrix}$$

Each has determinant $\pm 1$, hence non-singular.

Similarly for each of the $3$ choices of the row pattern and $3$ positions of that row. Thus total is indeed

$$36.$$

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8. Compare with options

The number of such matrices is $36$. Now evaluate the options:

• A: $5$ — false
• B: $6$ — false
• C: at least $7$ — true
• D: less than $4$ — false

Since this is an MCQ with single correct option, the only valid statement is:

$$\boxed{\text{C}}$$

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9. Comparison with stored correct answer

Stored correct answer is C. My derived answer is also C.

So they agree.