Let A be the set of all `2 xx 2` matrices whose determinant values is 1 and B be the set of all `2 xx 2` matrices whose determinant value is -1 and each entry of matrix is either 0 or 1, then

Options

A. `A subsetB`
B. n(B)=n(A)
C. `B subset A`
D. Nothing can be concluded

Answer

To solve the problem, we need to analyze the sets \( A \) and \( B \) defined as follows:

- Set \( A \): The set of all \( 2 \times 2 \) matrices whose determinant is 1.
- Set \( B \): The set of all \( 2 \times 2 \) matrices whose determinant is -1, where each entry of the matrix is either 0 or 1.

### Step 1: Determine the possible matrices in set \( A \)

A general \( 2 \times 2 \) matrix can be represented as:
\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
The determinant of this matrix is given by:
\[ \text{det}(A) = ad - bc \]
For the matrix to belong to set \( A \), we need:
\[ ad - bc = 1 \]

### Step 2: Analyze the conditions for \( ad = 1 \) and \( bc = 0 \)

Since \( a, b, c, d \) can only be 0 or 1, the only way for \( ad \) to equal 1 is if both \( a = 1 \) and \( d = 1 \). Thus, we have:
\[ ad = 1 \implies a = 1, d = 1 \]
Now we need \( bc = 0 \). This means at least one of \( b \) or \( c \) must be 0. The possible combinations for \( b \) and \( c \) are:
1. \( b = 0, c = 0 \)
2. \( b = 0, c = 1 \)
3. \( b = 1, c = 0 \)

This gives us the following matrices in set \( A \):
1. \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
2. \( \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \)
3. \( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \)

### Step 3: Determine the possible matrices in set \( B \)

For set \( B \), we again consider a general \( 2 \times 2 \) matrix:
\[ \begin{pmatrix} e & f \\ g & h \end{pmatrix} \]
The determinant is:
\[ \text{det}(B) = eh - fg \]
For the matrix to belong to set \( B \), we need:
\[ eh - fg = -1 \]

### Step 4: Analyze the conditions for \( eh = 0 \) and \( fg = 1 \)

For \( eh \) to equal 0, at least one of \( e \) or \( h \) must be 0. The only way for \( fg \) to equal 1 is if both \( f = 1 \) and \( g = 1 \). Thus, we have:
\[ fg = 1 \implies f = 1, g = 1 \]
Now we consider the possible values for \( e \) and \( h \):
1. \( e = 0, h = 0 \)
2. \( e = 0, h = 1 \)
3. \( e = 1, h = 0 \)

This gives us the following matrices in set \( B \):
1. \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \)
2. \( \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \)
3. \( \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \)

### Step 5: Compare sets \( A \) and \( B \)

Now, we have identified the matrices in both sets:
- Set \( A \):
1. \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
2. \( \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \)
3. \( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \)

- Set \( B \):
1. \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \)
2. \( \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \)
3. \( \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \)

### Conclusion

1. **Subset Check**: There are no common matrices between sets \( A \) and \( B \).
2. **Count of Elements**: Both sets \( A \) and \( B \) contain 3 matrices each.

Thus, we conclude that:
- \( A \) is not a subset of \( B \).
- \( B \) is not a subset of \( A \).
- The number of elements in \( A \) is equal to the number of elements in \( B \).

### Final Answer
The correct conclusion is that the number of elements in \( A \) is equal to the number of elements in \( B \).

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