The matrices which commule with `A=[{:(1,1),(0,1):}]` in case of multiplication
STATEMENT - 1 : Are always singular.
STATEMENT -2 : Are always non-singular.
STATEMENT -3 : Are always symmetric

Options

A. FFF
B. TTF
C. TTT
D. TFT

Answer

To determine which statements about matrices that commute with the matrix \( A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \) are true, we will analyze the conditions for a matrix \( B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) to commute with \( A \).

### Step-by-Step Solution:

1. **Define the Commutative Property**:
For two matrices \( A \) and \( B \) to commute, they must satisfy the equation \( AB = BA \).

2. **Calculate \( AB \)**:
\[ AB = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 \cdot a + 1 \cdot c & 1 \cdot b + 1 \cdot d \\ 0 \cdot a + 1 \cdot c & 0 \cdot b + 1 \cdot d \end{pmatrix} = \begin{pmatrix} a + c & b + d \\ c & d \end{pmatrix} \]

3. **Calculate \( BA \)**:
\[ BA = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a \cdot 1 + b \cdot 0 & a \cdot 1 + b \cdot 1 \\ c \cdot 1 + d \cdot 0 & c \cdot 1 + d \cdot 1 \end{pmatrix} = \begin{pmatrix} a & a + b \\ c & c + d \end{pmatrix} \]

4. **Set \( AB = BA \)**:
\[ \begin{pmatrix} a + c & b + d \\ c & d \end{pmatrix} = \begin{pmatrix} a & a + b \\ c & c + d \end{pmatrix} \]

5. **Equate the corresponding elements**:
- From the first row, first column:
\[ a + c = a \implies c = 0 \]
- From the first row, second column:
\[ b + d = a + b \implies d = a \]
- From the second row, first column:
\[ c = c \text{ (always true)} \]
- From the second row, second column:
\[ d = c + d \implies c = 0 \text{ (already established)} \]

6. **Conclusion about matrix \( B \)**:
Thus, the matrix \( B \) that commutes with \( A \) can be expressed as:
\[ B = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \]
where \( a \) can be any scalar and \( b \) can be any scalar.

7. **Determine the properties of \( B \)**:
- **Singularity**: The determinant of \( B \) is given by:
\[ \text{det}(B) = a \cdot a - 0 \cdot b = a^2 \]
- If \( a = 0 \), then \( B \) is singular.
- If \( a \neq 0 \), then \( B \) is non-singular.
- **Symmetry**: The matrix \( B \) is symmetric if \( B = B^T \):
\[ B^T = \begin{pmatrix} a & 0 \\ b & a \end{pmatrix} \]
- For \( B \) to be symmetric, \( b \) must equal \( 0 \). Therefore, \( B \) is not necessarily symmetric.

### Final Evaluation of Statements:
- **Statement 1**: "Are always singular." - **False** (can be non-singular if \( a \neq 0 \)).
- **Statement 2**: "Are always non-singular." - **False** (can be singular if \( a = 0 \)).
- **Statement 3**: "Are always symmetric." - **False** (only symmetric if \( b = 0 \)).